អនុវត្តន៍នៃផលគុណវ៉ិចទ័រ

៣. ដំណោះស្រាយលម្អិតនៃលំហាត់អនុវត្តន៍ (មេរៀនទី ៣)

ដំណោះស្រាយ

ក ដោយ \(ABCD\) ជារូបចតុមុខទៀងទាត់ \(A'\) ជាផ្ចិតរង្វង់ចារឹកក្រៅ \(\triangle BCD\)៖
\(A'B = R = \frac{a}{\sqrt{3}}\) (ដែល \(a=AB\))។
ក្នុង \(\triangle OA'B\) កែងត្រង់ \(A'\)៖ \(OA' = \frac{x}{3}\), \(OB = x\)៖
\(A'B^2 = OB^2 - OA'^2 = x^2 - \frac{x^2}{9} = \frac{8}{9}x^2\)
\(\implies a^2 = 3A'B^2 = \frac{8}{3}x^2\)។
ក្នុង \(\triangle OAB\)៖ \(\overrightarrow{OA}\cdot\overrightarrow{OB} = \frac{1}{2}(OA^2+OB^2-AB^2)\)
\(= \frac{1}{2}(2x^2 - \frac{8}{3}x^2) = \boxed{-\frac{1}{3}x^2}\)

ខ គណនា \(\cos\angle AOB\) និង \(AB\)៖
\(\bullet\ \cos\angle AOB = \frac{\overrightarrow{OA}\cdot\overrightarrow{OB}}{OA\cdot OB} = \frac{-\frac{1}{3}x^2}{x^2} = \boxed{-\frac{1}{3}}\)
\(\bullet\ AB = a = \sqrt{\frac{8}{3}x^2} = \boxed{\frac{2\sqrt{6}}{3}x\text{ (cm)}}\)

ដំណោះស្រាយ

ក រកវ៉ិចទ័រ \(\overrightarrow{AB}\) និង \(\overrightarrow{BC}\)៖
\(\bullet\ \overrightarrow{AB} = (2-2, 0-2, 1-2) = \boxed{(0, -2, -1)}\)
\(\bullet\ \overrightarrow{BC} = (0-2, -1-0, 3-1) = \boxed{(-2, -1, 2)}\)

ខ គណនាផលគុណស្កាលែ៖
\(\overrightarrow{AB}\cdot\overrightarrow{BC} = 0(-2) + (-2)(-1) + (-1)(2) = \boxed{0}\)
ដោយ \(\overrightarrow{AB}\cdot\overrightarrow{BC} = 0 \implies \overrightarrow{AB}\perp\overrightarrow{BC}\)
\(\implies \boxed{\triangle ABC \text{ ជាត្រីកោណកែងត្រង់ } B}\)

ដំណោះស្រាយ

ក សមីការប៉ារ៉ាម៉ែត្រនៃ \((D)\)៖
\(\boxed{(D)\colon \begin{cases} x = 1 + 2t \\ y = -3 + t \\ z = 2 + 3t \end{cases} \quad{} (t\in\mathbb{R})}\)

ខ ជំនួស \((D)\) ក្នុង \((Q)\)៖
\(2(1+2t)-(-3+t)+5(2+3t)+3=0\)
\(\iff 18t + 18 = 0 \implies t = -1\)
\(\implies \boxed{I(-1, -4, -1)}\)

ដំណោះស្រាយ

លំហាត់ ៤៖
\(\vec{u}\perp\vec{c} \iff \vec{u}\cdot\vec{c} = 0\)
\(\iff 3(x) + 1(1) + (-4)(-5) = 0\)
\(\iff 3x + 21 = 0 \implies \boxed{x = -7}\)

លំហាត់ ៥៖
\(\vec{a}\perp\vec{e} \iff \vec{a}\cdot\vec{e} = 0\)
\(\iff -2(1) + 1(y) + 7(4) = 0\)
\(\iff -2 + y + 28 = 0 \implies \boxed{y = -26}\)

ដំណោះស្រាយ

គណនាវ៉ិចទ័រ \(\overrightarrow{RP}\) និង \(\overrightarrow{RQ}\) ចេញពីកំពូល \(R\)៖
\(\bullet\ \overrightarrow{RP} = (1-(-1), 7-6, 2-1) = (2, 1, 1)\)
\(\bullet\ \overrightarrow{RQ} = (0-(-1), 7-6, -2-1) = (1, 1, -3)\)

ផលគុណស្កាលែ៖
\(\overrightarrow{RP}\cdot\overrightarrow{RQ} = 2(1) + 1(1) + 1(-3) = 2+1-3 = 0\)
\(\implies \overrightarrow{RP}\perp\overrightarrow{RQ} \implies \boxed{\triangle PQR \text{ ជាត្រីកោណកែងត្រង់ } R}\)

ដំណោះស្រាយ

ក \(\vec{u}(-1,1,1)\)៖
\(\|\vec{u}\| = \sqrt{1+1+1} = \boxed{\sqrt{3}}\)
\(\cos = \boxed{\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)}\)
ខ \(\vec{u}(3,0,0)\)៖
\(\|\vec{u}\| = \sqrt{9+0+0} = \boxed{3}\)
\(\cos = \boxed{(1, 0, 0)}\)
គ \(\vec{u}(1,1,1)\)៖
\(\|\vec{u}\| = \sqrt{1+1+1} = \boxed{\sqrt{3}}\)
\(\cos = \boxed{\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)}\)
ឃ \(\vec{u}(4,-1,0)\)៖
\(\|\vec{u}\| = \sqrt{16+1+0} = \boxed{\sqrt{17}}\)
\(\cos = \boxed{\left(\frac{4}{\sqrt{17}}, -\frac{1}{\sqrt{17}}, 0\right)}\)

ង \(\vec{u}(-1,-1,-1)\)៖
\(\|\vec{u}\| = \sqrt{1+1+1} = \boxed{\sqrt{3}}\)
\(\cos = \boxed{\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)}\)
ច \(\vec{u}(-3,-3,8)\)៖
\(\|\vec{u}\| = \sqrt{9+9+64} = \boxed{\sqrt{82}}\)
\(\cos = \boxed{\left(-\frac{3}{\sqrt{82}}, -\frac{3}{\sqrt{82}}, \frac{8}{\sqrt{82}}\right)}\)
ឆ \(\vec{u}(3,-1,-4)\)៖
\(\|\vec{u}\| = \sqrt{9+1+16} = \boxed{\sqrt{26}}\)
\(\cos = \boxed{\left(\frac{3}{\sqrt{26}}, -\frac{1}{\sqrt{26}}, -\frac{4}{\sqrt{26}}\right)}\)
ជ \(\vec{u}(-2,-3,-4)\)៖
\(\|\vec{u}\| = \sqrt{4+9+16} = \boxed{\sqrt{29}}\)
\(\cos = \boxed{\left(-\frac{2}{\sqrt{29}}, -\frac{3}{\sqrt{29}}, -\frac{4}{\sqrt{29}}\right)}\)

ដំណោះស្រាយ

ក \(\|\vec{u}\|=6, \|\vec{v}\|=4, \vec{u}\cdot\vec{v}=12\sqrt{2}\)៖
\(\cos\theta = \frac{12\sqrt{2}}{24} = \frac{\sqrt{2}}{2} \implies \boxed{\theta = 45^\circ = \frac{\pi}{4}}\)
ខ \(\|\vec{u}\|=3, \|\vec{v}\|=2, \vec{u}\cdot\vec{v}=-3\sqrt{2}\)៖
\(\cos\theta = \frac{-3\sqrt{2}}{6} = -\frac{\sqrt{2}}{2} \implies \boxed{\theta = 135^\circ = \frac{3\pi}{4}}\)

គ \(\|\vec{u}\|=3, \|\vec{v}\|=3, \vec{u}\cdot\vec{v}=-2-2+4=0\)៖
\(\cos\theta = 0 \implies \boxed{\theta = 90^\circ = \frac{\pi}{2}}\)
ឃ \(\|\vec{u}\|=3, \|\vec{v}\|=2, \vec{u}\cdot\vec{v}=-3\sqrt{2}\)៖
\(\cos\theta = -\frac{\sqrt{2}}{2} \implies \boxed{\theta = 135^\circ = \frac{3\pi}{4}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (-2, 0, 3), \; \overrightarrow{BC} = (3, 0, 2)\)
ខ \(AB = \sqrt{4+0+9} = \boxed{\sqrt{13}}\)
\(BC = \sqrt{9+0+4} = \boxed{\sqrt{13}}\)
គ \(\overrightarrow{AB}\cdot\overrightarrow{BC} = -6+0+6 = \boxed{0}\)

ឃ ដោយ \(\overrightarrow{BA}\cdot\overrightarrow{BC} = -\overrightarrow{AB}\cdot\overrightarrow{BC} = 0\)
\(\implies \cos\angle ABC = 0 \implies \boxed{\angle ABC = 90^\circ}\)
ដោយ \(AB = BC = \sqrt{13}\) និង \(\angle B = 90^\circ\)៖
\(\implies \boxed{\triangle ABC \text{ ជាត្រីកោណកែងសមបាតត្រង់ } B}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (2\sqrt{2}, -2, -2)\)
\(\overrightarrow{AC} = (2\sqrt{2}, 2, -2)\)
ខ \(AB = \sqrt{8+4+4} = \sqrt{16} = \boxed{4}\)
\(AC = \sqrt{8+4+4} = \sqrt{16} = \boxed{4}\)

គ \(\overrightarrow{AB}\cdot\overrightarrow{AC} = 8 - 4 + 4 = \boxed{8}\)
ឃ \(\cos\angle BAC = \frac{8}{4\times 4} = \frac{1}{2} \implies \boxed{\angle BAC = 60^\circ}\)
ដោយ \(AB = AC = 4\) និង \(\angle A = 60^\circ\)៖
\(\implies \boxed{\triangle ABC \text{ ជាត្រីកោណសម័ង្ស}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (-6, -1, -4)\)
\(\overrightarrow{AC} = (-1, 6, 4), \; \overrightarrow{BC} = (5, 7, 8)\)
ខ \(AB = \sqrt{36+1+16} = \boxed{\sqrt{53}}\)
\(AC = \sqrt{1+36+16} = \boxed{\sqrt{53}}\)
\(BC = \sqrt{25+49+64} = \boxed{\sqrt{138}}\)
គ ដោយ \(AB = AC = \sqrt{53} \implies \boxed{\triangle ABC \text{ សមបាតកំពូល } A}\)

ឃ \(ABD\) សមបាតកំពូល \(B \iff BD^2 = BA^2\)៖
\(\overrightarrow{BD} = (-4, 1, z+1) \implies BD^2 = 17+(z+1)^2\)
\(17 + (z+1)^2 = 53 \iff (z+1)^2 = 36\)
\(\implies z+1 = \pm 6 \implies \boxed{z = 5 \text{ ឬ } z = -7}\)
ង ទីប្រជុំទម្ងន់ \(G = \left(\frac{x_A+x_B+x_C}{3}, \dots\right)\)៖
\(G = \left(\frac{4-2+3}{3}, \frac{2+1+8}{3}, \frac{3-1+7}{3}\right) = \boxed{\left(\frac{5}{3}, \frac{11}{3}, 3\right)}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (3,-2,-1) \implies AB = \boxed{\sqrt{14}}\)
\(\overrightarrow{AC} = (1,-3,2) \implies AC = \boxed{\sqrt{14}}\)
\(\overrightarrow{BC} = (-2,-1,3) \implies BC = \boxed{\sqrt{14}}\)

ខ ដោយ \(AB = AC = BC = \sqrt{14}\)
\(\implies \boxed{\triangle ABC \text{ ជាត្រីកោណសម័ង្ស}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (-2,1,1) \implies AB = \boxed{\sqrt{6}}\)
\(\overrightarrow{AC} = (-2,-2,4) \implies AC = \boxed{2\sqrt{6}}\)

ខ \(\overrightarrow{AB}\cdot\overrightarrow{AC} = 4-2+4 = \boxed{6}\)
\(\cos\theta = \frac{6}{\sqrt{6}\cdot 2\sqrt{6}} = \frac{1}{2} \implies \boxed{\theta = 60^\circ = \frac{\pi}{3}}\)

ដំណោះស្រាយ

លំហាត់ ១៤៖
\(\overrightarrow{MN}=(0,4,-4), \overrightarrow{MP}=(-4,4,0) \implies MN=MP=4\sqrt{2}\)
\(\overrightarrow{MN}\cdot\overrightarrow{MP} = 16 \implies \cos\theta = \frac{16}{32} = \frac{1}{2} \implies \boxed{\theta = 60^\circ}\)
\(NP = 4\sqrt{2} \implies \boxed{\triangle MNP \text{ ជាត្រីកោណសម័ង្ស}}\)

លំហាត់ ១៥៖
\(\overrightarrow{AB}=(1,1,-1) \implies \overrightarrow{AB}\cdot\overrightarrow{AC} = -1+2-1 = \boxed{0}\)
\(\bullet\ C = A + \overrightarrow{AC} = \boxed{(0, 2, 2)}\)
\(\bullet\ M\) ជាចំណុចកណ្ដាល \(AC \implies \boxed{M\left(\frac{1}{2}, 1, \frac{3}{2}\right)}\)
\(\bullet\ G = \frac{A+B+C}{3} = \boxed{(1, 1, 1)}\)

ដំណោះស្រាយ

ក កាត់ \(A(-1,-1,4), \vec{u}=(1,1,-2)\)៖
\(\boxed{(L)\colon x = -1+t, \; y = -1+t, \; z = 4-2t}\)
ខ កាត់ \(B(1,3,-2), \vec{u}=\vec{n}_P=(1,-1,2)\)៖
\(\boxed{(L)\colon x = 1+t, \; y = 3-t, \; z = -2+2t}\)
គ កាត់ \(C(1,-1,2), \vec{u}=\vec{n}_P=(5,-1,3)\)៖
\(\boxed{(L)\colon x = 1+5t, \; y = -1-t, \; z = 2+3t}\)

ឃ កាត់ \(D(1,5,10), \vec{u}=\vec{k}=(0,0,1)\)៖
\(\boxed{(L)\colon x = 1, \; y = 5, \; z = 10+t}\)
ង កាត់ \(E(2,-1,0), \vec{u}=(2,-2,5)\)៖
\(\boxed{(L)\colon x = 2+2t, \; y = -1-2t, \; z = 5t}\)
ច កាត់ \(F(0,4,-2), \vec{u}=(5,-2,2)\)៖
\(\boxed{(L)\colon x = 5t, \; y = 4-2t, \; z = -2+2t}\)

ដំណោះស្រាយ

ក \(\boxed{(L)\colon x=2+t, \; y=-3+t, \; z=7-4t}\)
ខ \(\frac{x-3}{1/2} = \frac{y-4}{-1} = \frac{z-5}{1} \implies \boxed{\vec{u}_d = (1, -2, 2)}\)

គ \(\vec{v}\cdot\vec{u}_d = 1-2-8 = -9\)
\(\|\vec{v}\|=3\sqrt{2}, \|\vec{u}_d\|=3 \implies \cos\theta = \frac{9}{9\sqrt{2}} = \frac{\sqrt{2}}{2}\)
\(\implies \boxed{\theta = 45^\circ = \frac{\pi}{4}}\)

ដំណោះស្រាយ

ក \(2(x-2)+2(y-1)+1(z-3)=0\)
\(\iff \boxed{2x + 2y + z - 9 = 0}\)
ខ \(6(x+1)-3(y-3)-2(z-5)=0\)
\(\iff \boxed{6x - 3y - 2z + 25 = 0}\)
គ \(3(x-0)-7(y-0)+1(z-0)=0\)
\(\iff \boxed{3x - 7y + z = 0}\)

ឃ \(\vec{n}=\overrightarrow{PQ}=(1,3,-3)\)៖
\(1(x-4)+3(y+2)-3(z-1)=0 \iff \boxed{x+3y-3z+5=0}\)
ង ស្រប \((yOz) \implies \vec{n}=(1,0,0) \implies \boxed{x = -1}\)
ច ស្រប \((xOz) \implies \vec{n}=(0,1,0) \implies \boxed{y = 8}\)
ឆ ស្រប \((xOy)\) កាត់ \(O \implies \boxed{z = 0}\)

ដំណោះស្រាយ

ក \(3(x-1)-2(y-2)+5(z+3)=0\)
\(\iff \boxed{3x - 2y + 5z + 16 = 0}\)
ខ \(\vec{a}\cdot\vec{n} = 12-2-10 = 0 \implies \vec{a}\perp\vec{n}\)៖
\(\implies \boxed{\vec{a}(4,1,-2) \text{ ស្របនឹងប្លង់ } \alpha}\)

គ \(\vec{a}(15,-10,25) = 5\vec{n} \implies \vec{a}\parallel\vec{n}\)៖
\(\implies \boxed{\vec{a} \text{ កែងនឹងប្លង់ } \alpha}\)
ឃ វ៉ិចទ័រកែងនឹង \(\alpha\) គឺពហុគុណនៃ \(\vec{n}\)៖
\(\boxed{-\vec{n} = (-3, 2, -5)}\) (ឬ \(2\vec{n}=(6,-4,10)\))

ដំណោះស្រាយ

ក ជំនួសកូអរដោនេក្នុង \((\alpha)\)៖
\(\bullet\ 1+6-(-1)-8 = 0 \implies \boxed{A\in(\alpha)}\)
\(\bullet\ 3+10-1-8 = 4 \neq 0 \implies \boxed{B\notin(\alpha)}\)
\(\bullet\ 1+12-5-8 = 0 \implies \boxed{C\in(\alpha)}\)

ខ ប្រសព្វជាមួយអ័ក្សកូអរដោនេ៖
\(\bullet\ Ox\colon \boxed{(8,0,0)} \quad{} \bullet\ Oy\colon \boxed{(0,4,0)}\)
\(\bullet\ Oz\colon \boxed{(0,0,-8)}\)
គ វ៉ិចទ័រកែងនឹង \(\vec{n}(1,2,-1)\)៖
\(\boxed{\vec{u}=(2,-1,0)} \) និង \( \boxed{\vec{v}=(1,0,1)}\)

ដំណោះស្រាយ

រូបមន្ត៖ \(d = \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}\)
ក \(d = \frac{|21-12-6-8|}{\sqrt{49+36+36}} = \frac{5}{11} \implies \boxed{d = \frac{5}{11}}\)
ខ \(d = \frac{|7|}{\sqrt{4+1+4}} = \frac{7}{3} \implies \boxed{d = \frac{7}{3}}\)
គ \(d = \frac{|-18-8+12|}{\sqrt{81+1+16}} = \frac{14}{\sqrt{98}} = \boxed{\sqrt{2}}\)

ឃ \(d = \frac{|-1+2-6|}{\sqrt{1+1}} = \frac{5}{\sqrt{2}} = \boxed{\frac{5\sqrt{2}}{2}}\)
ង \(d = \frac{|21-1|}{\sqrt{4+49+1}} = \frac{20}{3\sqrt{6}} = \boxed{\frac{10\sqrt{6}}{9}}\)
ច \(d = \frac{|8-4-1-3|}{\sqrt{4+1+1}} = \frac{0}{\sqrt{6}} = \boxed{0}\) (\(R\in\alpha\))

ដំណោះស្រាយ

ក រូបមន្ត៖ \(d = \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}\)
\(d = \frac{|6-(-24)|}{\sqrt{64+1+16}} = \frac{30}{\sqrt{81}} = \boxed{\frac{10}{3}}\)

ខ \(d = \frac{|-12-(-7)|}{\sqrt{4+1+1}} = \frac{5}{\sqrt{6}} = \boxed{\frac{5\sqrt{6}}{6}}\)

ដំណោះស្រាយ

ក \(\cos\theta = \frac{|-2-4-3|}{\sqrt{14}\cdot 3} = \frac{3}{\sqrt{14}} \implies \boxed{\theta = \arccos\frac{3\sqrt{14}}{14}}\)
ខ \(\cos\theta = \frac{|-4|}{\sqrt{10}\cdot\sqrt{20}} = \frac{4}{10\sqrt{2}} \implies \boxed{\theta = \arccos\frac{\sqrt{2}}{5}}\)

គ \(\cos\theta = \frac{|12+8-42|}{9\cdot 7} = \frac{22}{63} \implies \boxed{\theta = \arccos\frac{22}{63}}\)
ឃ \(\cos\theta = \frac{|2|}{\sqrt{38}\cdot 1} = \frac{2}{\sqrt{38}} \implies \boxed{\theta = \arccos\frac{\sqrt{38}}{19}}\)

ដំណោះស្រាយ

ក \(\boxed{x^2+(y-1)^2+(z-2)^2 = 3}\)
ខ \(r^2 = IM^2 = 4+4+4=12\)៖
\(\boxed{x^2+(y+1)^2+(z-3)^2 = 12}\)
គ ផ្ចិត \(I(2,2,5), r^2 = 8\)៖
\(\boxed{(x-2)^2+(y-2)^2+(z-5)^2 = 8}\)

ឃ \(r = \frac{|-4+3-1|}{\sqrt{14}} = \frac{2}{\sqrt{14}} \implies r^2 = \frac{2}{7}\)៖
\(\boxed{(x-3)^2+(y+1)^2+(z-4)^2 = \frac{2}{7}}\)
ង \(r = \frac{|4+2+1-3|}{\sqrt{6}} = \frac{4}{\sqrt{6}} \implies r^2 = \frac{8}{3}\)៖
\(\boxed{(x-2)^2+(y-2)^2+(z+1)^2 = \frac{8}{3}}\)
ច \(S'\colon I'(1,-2,3), R'=3\)។ \(II'=7\)៖
ប៉ះក្រៅ \(r = 7-3=4 \implies \boxed{(x+5)^2+(y-1)^2+(z-1)^2 = 16}\)

ដំណោះស្រាយ

ក \((x-1)^2+y^2+(z+2)^2 = 9\)
\(\implies \boxed{I(1,0,-2), \; R = 3}\)
ខ \((x+1)^2+(y+7)^2+(z-5)^2 = 1\)
\(\implies \boxed{I(-1,-7,5), \; R = 1}\)
គ \((x+1)^2+(y-1)^2+z^2 = 9\)
\(\implies \boxed{I(-1,1,0), \; R = 3}\)

ឃ \((x-3)^2+(y+1)^2+(z-2)^2 = 33\)
\(\implies \boxed{I(3,-1,2), \; R = \sqrt{33}}\)
ង \((x-4)^2+(y+2)^2+(z+5)^2 = 36\)
\(\implies \boxed{I(4,-2,-5), \; R = 6}\)
ច \(x^2+y^2+z^2 = 4\)
\(\implies \boxed{I(0,0,0), \; R = 2}\)

ដំណោះស្រាយ

ក \(S\colon I(1,0,-2), R=3\) និង \(S'\colon I'(3,-1,2), R'=\sqrt{33}\)៖
\(II' = \sqrt{4+1+16} = \sqrt{21}\)។
ដោយ \(|R-R'| < II' < R+R'\)
\(\implies \boxed{S \text{ និង } S' \text{ កាត់គ្នាបង្កើតជារង្វង់}}\)

ខ \(S\colon I(4,-2,1), R=\sqrt{17}\) និង \(S'\colon I'(-2,1,2), R'=2\)៖
\(II' = \sqrt{36+9+1} = \sqrt{46} > R+R'\)
\(\implies \boxed{S \text{ និង } S' \text{ នៅក្រៅគ្នា (មិនជួបគ្នា)}}\)
គ \(S\colon I(-1,1,0), R=3\) និង \(S'\colon I'(1,0,-2), R'=3\)៖
\(II' = \sqrt{4+1+4} = 3 < R+R'=6\)
\(\implies \boxed{S \text{ និង } S' \text{ កាត់គ្នាបង្កើតជារង្វង់}}\)

ដំណោះស្រាយ

ក \(S\colon I(-1,4,2), R=3\)៖
\(d(I,P) = \frac{|-2-8+2-1|}{3} = 3 = R\)
\(\implies \boxed{\text{ប្លង់ } (P) \text{ ប៉ះស្វ៊ែ } S}\)
ខ \(S\colon I(1,-3,-2), R=6\)៖
\(d(I,P) = \frac{|1-6+4-5|}{3} = 2 < R\)
\(\implies \boxed{\text{ប្លង់ } (P) \text{ កាត់ស្វ៊ែ } S \text{ តាមរង្វង់}}\)

គ \(S\colon I(-1,2,1), R=2\)៖
\(d(I,P) = \frac{|-1+2-2-11|}{\sqrt{6}} = 2\sqrt{6} > R\)
\(\implies \boxed{\text{ប្លង់ } (P) \text{ មិនជួបស្វ៊ែ } S}\)

ដំណោះស្រាយ

លំហាត់ ២៨៖
\(\bullet\ \vec{i}\times\vec{j}=\boxed{\vec{k}}, \; \vec{j}\times\vec{k}=\boxed{\vec{i}}, \; \vec{k}\times\vec{i}=\boxed{\vec{j}}\)
\(\bullet\ \vec{j}\times\vec{i}=\boxed{-\vec{k}}, \; \vec{k}\times\vec{j}=\boxed{-\vec{i}}, \; \vec{i}\times\vec{k}=\boxed{-\vec{j}}\)
លំហាត់ ២៩៖
ក \((2,-3,1)\times(1,-2,1) = \boxed{(-1, -1, -1)}\)
ខ \((12,-3,0)\times(-2,5,0) = \boxed{(0, 0, 54)}\)
គ \((1,1,1)\times(2,1,-1) = \boxed{(-2, 3, -1)}\)

លំហាត់ ៣០៖
ក \(\vec{x} = 3\vec{v}-2\vec{u}\)៖
\(= 3(7,9,-8)-2(4,-2,5)\\ = (21-8, 27+4, -24-10)\\ = \boxed{(13, 31, -34)}\)
ខ \(2\vec{x} = 4(\vec{u}-\vec{v}) \implies \vec{x} = 2(\vec{u}-\vec{v})\)៖
\(= 2(-3, -11, 13) = \boxed{(-6, -22, 26)}\)

ដំណោះស្រាយ

ក \(\vec{u}\parallel\vec{v} \iff \frac{3x}{2} = \frac{5}{-1} = \frac{4y-2}{5} = -5\)៖
\(\bullet\ \frac{3x}{2} = -5 \implies \boxed{x = -\frac{10}{3}}\)
\(\bullet\ \frac{4y-2}{5} = -5 \implies 4y = -23 \implies \boxed{y = -\frac{23}{4}}\)

ខ \(\vec{u}\perp\vec{w} \iff \vec{u}\cdot\vec{w} = 0\)៖
\(2(z-1) - 1(2) + 5(z+1) = 0\)
\(\iff 7z + 1 = 0 \implies \boxed{z = -\frac{1}{7}}\)

ដំណោះស្រាយ

លំហាត់ ៣២៖
\(\overrightarrow{AB}=(2,-3,6) \implies AB = 7\)
\(\overrightarrow{AC}=(3,6,2) \implies AC = 7\)
\(\overrightarrow{BC}=(1,9,-4) \implies BC = \sqrt{98}\)
\(\overrightarrow{AB}\cdot\overrightarrow{AC} = 6-18+12 = 0 \implies \angle A = 90^\circ\)
\(\implies \boxed{\triangle ABC \text{ ជាត្រីកោណកែងសមបាតត្រង់ } A}\)

លំហាត់ ៣៣៖
\(\overrightarrow{AB}=(-x, y-2, 2), \; \overrightarrow{AC}=(3-x, 2, -1)\)
\(\bullet\ \overrightarrow{AB}\cdot\overrightarrow{AC} = 0 \iff x^2-3x+2y-6=0\)
\(\bullet\ AB^2=AC^2 \iff (y-2)^2+6x-10=0\)
ដោះស្រាយប្រព័ន្ធ \(\implies \boxed{x = 1, \; y = 4}\)

ដំណោះស្រាយ

ក គណនាវ៉ិចទ័រ៖
\(\bullet\ \vec{u} = \vec{a}+\vec{b}+\vec{c} = \boxed{(2, 2, 3)}\)
\(\bullet\ \vec{v} = 2\vec{a}-\vec{b}+3\vec{c} = \boxed{(9, 4, -4)}\)
\(\bullet\ \vec{w} = 2\vec{u}+3\vec{v} = \boxed{(31, 16, -6)}\)

ខ ប្រព័ន្ធសមីការ៖
\(\begin{cases} \alpha - \beta + 2\gamma = 1 \\ -2\alpha + \beta + 3\gamma = 3 \\ \alpha + 3\beta - \gamma = 6 \end{cases} \implies \boxed{\alpha = 1, \; \beta = 2, \; \gamma = 1}\)

ដំណោះស្រាយ

ក \(\vec{u}\times\vec{v} = (12-40, -(16+5), 32+3)\)
\(= \boxed{(-28, -21, 35)}\)
ខ \(\vec{u}\times\vec{v} = (10+10, -(-15+55), -6-22)\)
\(= \boxed{(20, -40, -28)}\)

គ \(\vec{u}\times\vec{v} = (12+48, -(16-40), 96+60)\)
\(= \boxed{(60, 24, 156)}\)
ឃ \(\vec{u}\times\vec{v} = (21-3, -(-7-0), -3-0)\)
\(= \boxed{(18, 7, -3)}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = \boxed{(1, 2, 2)}\)
\(\overrightarrow{BC} = \boxed{(2, -3, -6)}, \; \overrightarrow{AC} = \boxed{(3, -1, -4)}\)
ខ \(AB = \sqrt{1+4+4} = \boxed{3}\)
\(BC = \sqrt{4+9+36} = \boxed{7}\)
\(AC = \sqrt{9+1+16} = \boxed{\sqrt{26}}\)

គ \(ABCD\) ជាប្រលេឡូក្រាម \(\iff \overrightarrow{AD} = \overrightarrow{BC}\)៖
\(D = A + \overrightarrow{BC} = (1+2, -1-3, 2-6)\)
\(\implies \boxed{D(3, -4, -4)}\)

ដំណោះស្រាយ

ក \(\vec{a}\cdot\vec{b} = -1-8+16 = \boxed{7}\)
ខ \(\|\vec{a}\| = \sqrt{1+4+4} = \boxed{3}\)
\(\|\vec{b}\| = \sqrt{1+16+64} = \boxed{9}\)
\(\cos(\vec{a},\vec{b}) = \frac{7}{3\times 9} = \boxed{\frac{7}{27}}\)

គ \(\bullet\ \vec{a}\cdot\vec{c} = \vec{b}\cdot\vec{c} \iff x-3y = 12\)
\(\bullet\ (\vec{a}+\vec{b})\cdot\vec{c} = 52 \iff (0,2,10)\cdot(x,y,4)=52\)
\(\iff 2y + 40 = 52 \implies \boxed{y = 6}\)
\(x = 12 + 3(6) \implies \boxed{x = 30}\)

ដំណោះស្រាយ

ក \(\overrightarrow{BC}=(2,4,4)=2(1,2,2)\)៖
\(\bullet\ (P)\colon 1(x+1)+2(y-2)+2(z-3)=0 \iff \boxed{x+2y+2z-9=0}\)
\(\bullet\ (BC)\colon \boxed{x = t, \; y = -1+2t, \; z = 1+2t}\)

ខ ជំនួស \((BC)\) ក្នុង \((P)\)៖ \(9t-9=0 \implies t=1\)
\(\implies \boxed{H(1, 1, 3)}\)
គ \(H\) ជាចំណុចកណ្ដាល \(BC\) និង \(AH\perp BC\)៖
\(\implies \boxed{\triangle ABC \text{ ជាត្រីកោណសមបាតកំពូល } A}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}=\overrightarrow{DC}=(-4,-2,0) \implies \boxed{ABCD \text{ ជាប្រលេឡូក្រាម}}\)
ខ \((BC)\colon x=0, y=-2+t, z=t\)
\((ABC)\colon \boxed{x - 2y + 2z - 4 = 0}\)
គ \(H(0,-1,1) \implies AH=3\sqrt{2}\)
\(S_{ABCD} = BC\cdot AH = 2\sqrt{2}\cdot 3\sqrt{2} = \boxed{12}\)

ឃ \(\overrightarrow{AS}=(-3,6,-6)=-3(1,-2,2) \parallel \vec{n}_{ABC}\)
\(\implies (SA)\perp (ABCD)\) មានកម្ពស់ \(h=SA=9\)៖
\(\boxed{V = \frac{1}{3}S_{ABCD}\cdot h = \frac{1}{3}(12)(9) = 36}\)

ដំណោះស្រាយ

លំហាត់ ៤០៖
\(\bullet\ H = (\Delta)\cap(\alpha) \implies \boxed{H(2, 4, 1)}\)
\(\bullet\ A' = 2H - A = \boxed{(3, 6, -1)}\)
\(\bullet\ (\beta)\colon \boxed{x + 2y - 2z - 17 = 0}\)

លំហាត់ ៤១៖
\(\bullet\ A = (\Delta)\cap(\alpha) \implies \boxed{A(1, 2, 3)}\)
\(\bullet\ \sin\theta = \frac{9}{3\sqrt{2}\cdot 3} = \frac{\sqrt{2}}{2} \implies \boxed{\theta = 45^\circ}\)
\(\bullet\ (\beta)\colon \boxed{x + 4y - z - 6 = 0}\)

ដំណោះស្រាយ

ក \((S)\colon I(-1,4,2), R=3\)៖
\(d(I,\alpha) = \frac{|-2-8+2-1|}{3} = 3 = R \implies \boxed{(\alpha) \text{ ប៉ះស្វ៊ែ } S}\)
ចំណុចប៉ះ \(\boxed{A(1, 2, 3)}\)

ខ \((\Delta)\colon \boxed{x=1+2t, \; y=2-2t, \; z=3+t}\)
\(B\) ជាចំណុចឆ្លុះនៃ \(A\) ធៀបនឹង \(I\)៖
\(B = 2I - A = \boxed{(-3, 6, 1)}\)
ប្លង់ \((\beta)\) ប៉ះស្វ៊ែត្រង់ \(B\)៖ \(\boxed{2x - 2y + z + 17 = 0}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (-3, 4, 2)\)
\(\overrightarrow{AC} = (-2, 1, 0), \; \overrightarrow{AD} = (-2, 0, -1)\)
ខ \(\overrightarrow{AB}\times\overrightarrow{AC} = (-2, -4, 5)\)
\((\overrightarrow{AB}\times\overrightarrow{AC})\cdot\overrightarrow{AD} = 4+0-5 = -1\)
\(\implies \boxed{V = \frac{1}{6}|-1| = \frac{1}{6}\text{ (ឯកតាមាឌ)}}\)

គ \(\overrightarrow{BC}=(1,-3,-2), \overrightarrow{BD}=(1,-4,-3)\)
\(\implies \overrightarrow{BC}\times\overrightarrow{BD} = (1, 1, -1)\)
\((BCD)\colon 1(x-1)+1(y-2)-1(z-2)=0\)
\(\iff \boxed{x + y - z - 1 = 0}\)

ដំណោះស្រាយ

ក \(\boxed{(D)\colon x=t, \; y=t, \; z=t \quad{} (t\in\mathbb{R})}\)
ខ \(\overrightarrow{AB}=(-1,4,-1), \overrightarrow{AC}=(-3,3,2)\)
\(\implies \overrightarrow{AB}\times\overrightarrow{AC} = \boxed{(11, 5, 9)}\)
\((ABC)\colon 11(x-3)+5y+9(z-1)=0\)
\(\iff \boxed{11x + 5y + 9z - 42 = 0}\)

គ ជំនួស \((D)\) ក្នុង \((ABC)\)៖
\(11t + 5t + 9t - 42 = 0 \iff 25t = 42\)
\(\implies t = \frac{42}{25}\)
\(\implies \boxed{I\left(\frac{42}{25}, \frac{42}{25}, \frac{42}{25}\right)}\)

ដំណោះស្រាយ

លំហាត់ ៤៥៖
\(\bullet\ \overrightarrow{AB} = (-1,-3,0) \implies \boxed{AB = \sqrt{10}}\)
\(\bullet\ (AB)\colon \boxed{x = 1-t, \; y = 1-3t, \; z = 1}\)
\(\bullet\ I(1/2, -1/2, 1), R^2 = 5/2\)៖
\(\boxed{(x-1/2)^2+(y+1/2)^2+(z-1)^2 = 5/2}\)

លំហាត់ ៤៦៖
\(\bullet\ \overrightarrow{AB}\times\overrightarrow{AC} = \boxed{(6, 5, 2)}\)
\(\bullet\ S_{ABC} = \frac{1}{2}\sqrt{36+25+4} = \boxed{\frac{\sqrt{65}}{2}}\)
\(\bullet\ (ABC)\colon \boxed{6x + 5y + 2z - 16 = 0}\)
\(\bullet\ d(O) = \frac{16}{\sqrt{65}} \implies \boxed{V = \frac{8}{3}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AK} = (-3,-3,3) \implies R^2 = AK^2 = 9+9+9 = 27\)៖
\(\boxed{(S)\colon (x-3)^2 + (y-2)^2 + z^2 = 27}\)

ខ \(\vec{n} = \overrightarrow{AK} = -3(1,1,-1)\)៖
\((P)\colon 1(x-0)+1(y+1)-1(z-3)=0 \iff \boxed{x + y - z + 4 = 0}\)

ដំណោះស្រាយ

លំហាត់ ៤៨៖
\(\bullet\ (P)\colon 1(x-0)-1(y-2)-1(z-2)=0 \iff \boxed{x-y-z+4=0}\)
\(\bullet\ H = (P)\cap(D) \implies 3t+2=0 \implies \boxed{H\left(\frac{1}{3}, \frac{5}{3}, \frac{8}{3}\right)}\)
\(\bullet\ d(A, D) = AH = \boxed{\frac{\sqrt{6}}{3}}\)

លំហាត់ ៤៩៖
\(\bullet\ \overrightarrow{AB}\times\overrightarrow{AC} = \boxed{(9, 3, 3) = 3(3, 1, 1)}\)
\(\bullet\ (ABC)\colon \boxed{3x + y + z - 4 = 0}\)
\(\bullet\ d(O) = \frac{4}{\sqrt{11}}, S = \frac{3\sqrt{11}}{2} \implies \boxed{V = 2}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}=(-2,1,2), \overrightarrow{AC}=(-1,2,-1)\)
\(\implies \overrightarrow{AB}\times\overrightarrow{AC} = \boxed{(-5, -4, -3)}\)
\(S_{ABC} = \frac{1}{2}\sqrt{25+16+9} = \boxed{\frac{5\sqrt{2}}{2}}\)

ខ \((ABC)\colon \boxed{5x + 4y + 3z - 13 = 0}\)
គ \(d(D, ABC) = \frac{|5+4+3-13|}{\sqrt{50}} = \boxed{\frac{\sqrt{2}}{10}}\)
\(V_{DABC} = \frac{1}{3}S_{ABC}\cdot d = \frac{1}{3}\frac{5\sqrt{2}}{2}\frac{\sqrt{2}}{10} = \boxed{\frac{1}{6}}\)

ដំណោះស្រាយ

លំហាត់ ៥១៖
\(\bullet\ \overrightarrow{AB}\times\overrightarrow{AC} = (10, 3, 2) \implies \boxed{(ABC)\colon 10x+3y+2z-6=0}\)
\(\bullet\ d(O) = \frac{6}{\sqrt{100+9+4}} = \boxed{\frac{6}{\sqrt{113}}}\)
\(\bullet\ V_{OABC} = \frac{1}{6}|1(2)(3)| = \boxed{1}\)

លំហាត់ ៥២៖
\(\bullet\ B = A + \overrightarrow{AB} = \boxed{(0, 2, 2)}\)
\(\bullet\ (AB)\colon \boxed{x = 1-t, \; y = 2t, \; z = 1+t}\)
\(\bullet\ (S)\colon \boxed{(x-1)^2 + y^2 + (z-1)^2 = 6}\)
\(\bullet\ \) ប្លង់ប៉ះត្រង់ \( B\colon \boxed{x - 2y - z + 6 = 0}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}=(-2,1,2), \overrightarrow{AC}=(-2,3,1)\)
\(\implies \overrightarrow{AB}\times\overrightarrow{AC} = (-5,-2,-4) = -(5,2,4)\)
\((ABC)\colon 5(x-2)+2(y-0)+4(z-1)=0\)
\(\iff \boxed{5x + 2y + 4z - 14 = 0}\)

ខ ស្វ៊ែអង្កត់ផ្ចិត \(AB\)៖
\(\bullet\) ផ្ចិត \(I = \frac{A+B}{2} = \left(1, \frac{1}{2}, 2\right)\)
\(\bullet\) កាំ \(R = \frac{AB}{2} = \frac{3}{2} \implies R^2 = \frac{9}{4}\)
\(\implies \boxed{(x-1)^2 + \left(y-\frac{1}{2}\right)^2 + (z-2)^2 = \frac{9}{4}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}=(1,-1,2), \overrightarrow{AC}=(-2,1,-1), \overrightarrow{AD}=(1,3,1)\)៖
\(\bullet\ \overrightarrow{AB}\cdot\overrightarrow{AD} = 1-3+2 = 0 \implies AB\perp AD\)
\(\bullet\ \overrightarrow{AC}\cdot\overrightarrow{AD} = -2+3-1 = 0 \implies AC\perp AD\)
\(\implies AD\perp (ABC) \implies \boxed{DA \text{ ជាកម្ពស់តេត្រាអ៊ែត}}\)

ខ \(\bullet\ DA = \sqrt{1+9+1} = \boxed{\sqrt{11}}\)
\(\bullet\ \overrightarrow{AB}\times\overrightarrow{AC} = (-1, -3, -1) \implies S_{ABC} = \frac{\sqrt{11}}{2}\)
\(\bullet\ V = \frac{1}{3}S_{ABC}\cdot DA = \frac{1}{3}\frac{\sqrt{11}}{2}\sqrt{11} = \boxed{\frac{11}{6}\text{ (ឯកតាមាឌ)}}\)

ដំណោះស្រាយ

ក \(\bullet\ (D)\colon \boxed{x=t, \; y=t, \; z=3+t}\)
\(\bullet\ \overrightarrow{SM}\perp\vec{u} \iff x+y+z-3=0\)៖
\(\implies \boxed{\text{សំណុំចំណុច } M \text{ ជាប្លង់ } x+y+z-3=0}\)

ខ \(SA = SB = AB = 3\sqrt{2} \implies \boxed{\triangle SAB \text{ សម័ង្ស}}\)
\(\bullet\ \overrightarrow{SA}\times\overrightarrow{SB} = (9,9,9) \implies \boxed{S_{SAB} = \frac{9\sqrt{3}}{2}}\)
\(\bullet\ \) ស្វ៊ែផ្ចិត \( O\colon \boxed{x^2 + y^2 + z^2 = 9}\)

ដំណោះស្រាយ

ក \(\bullet\ \overrightarrow{AM}\cdot\overrightarrow{BM} = 0 \implies \boxed{\text{ស្វ៊ែអង្កត់ផ្ចិត } AB}\)
\((x+1)(x-2) + y(y-1) + z(z-1) = 0\)
\(\bullet\ \overrightarrow{BC}\times\overrightarrow{BD} = \boxed{(-7, -5, 3)}\)
\(\implies S_{BCD} = \frac{1}{2}\sqrt{49+25+9} = \boxed{\frac{\sqrt{83}}{2}}\)

ខ \(\bullet\ (BCD)\colon \boxed{7x + 5y - 3z - 11 = 0}\)
\(\bullet\ N = (BCD)\cap(d) \implies \boxed{N(-8, 8, -9)}\)
\(\bullet\ d(A, BCD) = \frac{13}{\sqrt{83}}\)
\(\implies \boxed{V = \frac{1}{3}\frac{\sqrt{83}}{2}\frac{13}{\sqrt{83}} = \frac{13}{6}}\)

ដំណោះស្រាយ

ក \(z=t \implies \boxed{(D)\colon x=2t-12, \; y=12-3t, \; z=t}\)
\(\vec{u} = (2,-3,1)\)
\(\vec{p}\times\vec{q} = (3,4,6)\times(1,1,1) = (-2,3,-1) = -\vec{u} \parallel \vec{u}\)

ខ \(A(4,0,0), B(0,3,0), C(0,0,2)\)
\(\bullet\ \overrightarrow{AB}\times\overrightarrow{AC} = (6,8,12) \implies \boxed{S_{ABC} = \sqrt{61}}\)
\(\bullet\ d(O, P) = \frac{12}{\sqrt{61}} \implies \boxed{V_{OABC} = 4}\)

ដំណោះស្រាយ

ក \(\bullet\ (S)\colon \boxed{(x-1)^2 + y^2 + (z-1)^2 = 2}\)
\(\bullet\ \vec{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \boxed{(6, 0, 6) = 6(1,0,1)}\)
\(\bullet\ S_{ABC} = \frac{1}{2}\sqrt{36+36} = \boxed{3\sqrt{2}}\)

ខ \(\bullet\ \overrightarrow{AB}\cdot\overrightarrow{AC} = 0 \implies \boxed{\triangle ABC \text{ កែងត្រង់ } A}\)
\(\bullet\ (P)\colon 1(x-0)+1(z-2)=0 \iff \boxed{x + z - 2 = 0}\)
\(\bullet\ 2t - t - 2 = 0 \implies t = 2 \implies \boxed{M(4, 7, -2)}\)

ដំណោះស្រាយ

ក \(\bullet\ (S)\colon \boxed{(x-1)^2 + (y-2)^2 + (z-1)^2 = 2}\)
\(\bullet\ \vec{n} = \boxed{(4,0,4) = 4(1,0,1)}\)
\(\bullet\ S_{ABC} = \boxed{2\sqrt{2}}\)
\(\bullet\ (ABC)\colon \boxed{x + z - 2 = 0}\)

ខ \(\bullet\ M(2,0,0), N(0,0,2)\)
\(\overrightarrow{AB}=\overrightarrow{MN}=(-2,0,2)\) និង \(\overrightarrow{MA}\cdot\overrightarrow{MN}=0\)
\(\implies \boxed{ABMN \text{ ជាចតុកោណកែង}}\)
\(\bullet\ d(D, ABC) = \sqrt{2} \implies \boxed{V = \frac{4}{3}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}=(1,1,0), \overrightarrow{AC}=(-1,1,0), \overrightarrow{AD}=(0,0,\sqrt{2})\)
\(AB=AC=AD=\sqrt{2}\) និងកែងគ្នាពីរៗ។
\(\implies \boxed{V = \frac{1}{6}(\sqrt{2})^3 = \frac{\sqrt{2}}{3}}\)

ខ \(\bullet\ \vec{N} = \overrightarrow{BC}\times\overrightarrow{BD} = \boxed{(0, 2\sqrt{2}, 2)}\)
\(\bullet\ S_{BCD} = \frac{1}{2}\sqrt{8+4} = \boxed{\sqrt{3}}\)
\(\bullet\ (BCD)\colon \boxed{\sqrt{2}y + z - 3\sqrt{2} + 2 = 0}\)
\(\bullet\ \boxed{H\left(1, \frac{8}{3}, -2+\frac{\sqrt{2}}{3}\right)}\)

ដំណោះស្រាយ

ក \(D(0,0,0), A(1,0,0), C(0,3,0), B(1,3,0)\)
\(H(0,0,2), E(1,0,2), G(0,3,2), F(1,3,2)\)

ខ \(\frac{x}{1}+\frac{y}{3}+\frac{z}{2}=1 \iff \boxed{6x+2y+3z-6=0}\)
\(d(D, ACH) = \frac{6}{\sqrt{36+4+9}} = \boxed{\frac{6}{7}}\)

ដំណោះស្រាយ

លំហាត់ ៦២៖
\(\bullet\ \overrightarrow{AB}\times\overrightarrow{AC} = 4(-5, 3, -10)\)៖
\((ABC)\colon \boxed{5x - 3y + 10z = 0}\)
\(\bullet\ 5(5)-3(3)+10(-1) = 6 \neq 0 \implies \boxed{D\notin(ABC)}\)

លំហាត់ ៦៣៖
\(\bullet\ AB=BC=AC=\sqrt{2} \implies \boxed{\triangle ABC \text{ សម័ង្ស}}\)
\(\bullet\ \overrightarrow{AB}\times\overrightarrow{AC} = \boxed{(1, 1, 1)}\)
\(\bullet\ (S)\colon \boxed{(x-1/2)^2 + y^2 + (z-1/2)^2 = 1/2}\)
\(\bullet\ \) ប្លង់ប៉ះត្រង់ \( C\colon \boxed{x - z + 1 = 0}\)

ដំណោះស្រាយ

ក \(AB = 4, AC = 4, BC = 4\)
\(\cos\angle BAC = \frac{1}{2} \implies \angle BAC = 60^\circ\)
\(\implies \boxed{\triangle ABC \text{ ជាត្រីកោណសម័ង្ស}}\)

ខ \(\overrightarrow{AB}\times\overrightarrow{AC} = \boxed{8\sqrt{2}(1, -1, 1)}\)
\(S_{ABC} = \frac{\sqrt{3}}{4}(4)^2 = \boxed{4\sqrt{3}\text{ (ឯកតាផ្ទៃ)}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{MN}=(1,1,1), \overrightarrow{MP}=(2,2,-2)\)
\(\implies \overrightarrow{MN}\times\overrightarrow{MP} = \boxed{(-4, 4, 0)}\)
\(S_{MNP} = \frac{1}{2}\sqrt{16+16} = \boxed{2\sqrt{2}}\)

ខ \((MNP)\colon -(x+1)+y=0 \iff \boxed{x - y + 1 = 0}\)
\(\bullet\ d(O, MNP) = \frac{1}{\sqrt{2}} = \boxed{\frac{\sqrt{2}}{2}}\)
\(\bullet\ V_{OMNP} = \frac{1}{3}(2\sqrt{2})\left(\frac{\sqrt{2}}{2}\right) = \boxed{\frac{2}{3}}\)

ដំណោះស្រាយ

ក \((x-1)^2+(y-2)^2+(z+2)^2 = 9\)
\(\implies \boxed{I(1, 2, -2), \; R = 3}\)

ខ \(d(I, P) = \frac{|1+4-4+5|}{3} = 2 < R=3\)
\(\implies \boxed{\text{ប្លង់ } (P) \text{ កាត់ស្វ៊ែ } (S) \text{ តាមរង្វង់}}\)
កាំរង្វង់ប្រសព្វ៖ \(r = \sqrt{R^2-d^2} = \sqrt{9-4} = \boxed{\sqrt{5}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}\times\overrightarrow{AC} = \boxed{(3, 10, -2)}\)
\((ABC)\colon \boxed{3x + 10y - 2z + 1 = 0}\)
ខ \(D = B+\overrightarrow{AC} = \boxed{(3, 0, 5)}\)
\(S_{ABDC} = \sqrt{9+100+4} = \boxed{\sqrt{113}}\)

គ \(d(F, ABC) = \frac{|9+100+4+1|}{\sqrt{113}} = \boxed{\frac{114}{\sqrt{113}}}\)
\(S_{ABC} = \frac{\sqrt{113}}{2}\)
\(V = \frac{1}{3}\frac{\sqrt{113}}{2}\frac{114}{\sqrt{113}} = \boxed{19}\)

ដំណោះស្រាយ

ក \(\overrightarrow{BA}\cdot\overrightarrow{BC} = 0+2-2 = \boxed{0} \implies \boxed{\triangle ABC \text{ កែងត្រង់ } B}\)
ខ \((L)\colon \boxed{x = 2+2t, \; y = 2+t, \; z = 2-2t}\)

គ \((P)\colon 2(x-4)+1(y-1)-2(z+1)=0 \iff \boxed{2x+y-2z-11=0}\)
ប្រសព្វ \(9t-9=0 \implies t=1 \implies \boxed{(4, 3, 0)}\)
ឃ \(\boxed{(S)\colon (x-1)^2 + (y-1)^2 + (z-1)^2 = 10}\)

ដំណោះស្រាយ

ក \(\overrightarrow{BA}\times\overrightarrow{BC} = \boxed{(-5, 2, 2)}\)
\((\alpha)\colon \boxed{5x - 2y - 2z - 1 = 0}\)
ខ \(S_{ABC} = \frac{\sqrt{33}}{2}, \; d(O) = \frac{1}{\sqrt{33}}\)
\(\implies \boxed{V_{OABC} = \frac{1}{6}}\)

គ \(MA^2-MB^2=2\)៖
\((x-1)^2+(y-1)^2+(z-1)^2 - \left[(x+1)^2+(y+2)^2+(z+1)^2\right] = 2\)
\(\iff -4x - 6y - 4z - 3 = 2\)
\(\iff \boxed{4x + 6y + 4z + 5 = 0 \text{ (ប្លង់)}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}\times\overrightarrow{AC} = 4(-5, 3, -10)\)
\(S_{ABC} = \frac{1}{2}\sqrt{400+144+1600} = \boxed{2\sqrt{134}}\)
ខ \((P)\colon \boxed{5x - 3y + 10z = 0}\)

គ \((L)\colon \boxed{x = 5+5t, \; y = 3-3t, \; z = -1+10t}\)
ឃ \(R = d(D, P) = \frac{|25-9-10|}{\sqrt{134}} = \frac{6}{\sqrt{134}} \implies R^2 = \frac{18}{67}\)៖
\(\boxed{(S)\colon (x-5)^2 + (y-3)^2 + (z+1)^2 = \frac{18}{67}}\)

ដំណោះស្រាយ

ក \(\vec{n} = \overrightarrow{CA}\times\overrightarrow{CB} = \boxed{(-6, -3, -2)}\)
\(S_{ABC} = \frac{1}{2}\sqrt{36+9+4} = \boxed{\frac{7}{2}}\)
ខ ប្លង់កាត់ \(O\colon \boxed{6x + 3y + 2z = 0}\)

គ \(\bullet\ V_{OABC} = \frac{1}{6}|x_A y_B z_C| = \frac{1}{6}(2)(1)(3) = \boxed{1}\)
\(\bullet\ d(O, ABC) = \frac{3V}{S} = \frac{3(1)}{7/2} = \boxed{\frac{6}{7}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}\times\overrightarrow{AC} = \boxed{(4, 7, -1)}\)
\((ABC)\colon \boxed{4x + 7y - z - 6 = 0}\)
ខ \(O(0,0,0)\colon -6\neq 0 \implies \boxed{O\notin(ABC)}\)
\(d(O) = \frac{6}{\sqrt{66}}, S = \frac{\sqrt{66}}{2} \implies \boxed{V = 1}\)

គ \((L)\colon \boxed{x = 1+4t, \; y = 1+7t, \; z = 2-t}\)
ស្វ៊ែអង្កត់ផ្ចិត \(AB\)៖ ផ្ចិត \(I(1, 1/2, 3/2), R^2 = 6/4 = 3/2\)៖
\(\boxed{(x-1)^2 + \left(y-\frac{1}{2}\right)^2 + \left(z-\frac{3}{2}\right)^2 = \frac{3}{2}}\)

ដំណោះស្រាយ

\(\bullet\ (ABC)\colon \frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1 \iff \boxed{6x+3y+2z-6=0}\)
\(\bullet\ S_{ABC} = \frac{1}{2}\sqrt{36+9+4} = \boxed{\frac{7}{2}}\)
\(\bullet\ OH = \frac{6}{\sqrt{49}} = \boxed{\frac{6}{7}}\)

\(\bullet\ \frac{1}{OH^2} = \frac{49}{36} = 1 + \frac{1}{4} + \frac{1}{9} = \frac{1}{OA^2}+\frac{1}{OB^2}+\frac{1}{OC^2}\)
\(\bullet\ \boxed{V_{OABC} = \frac{1}{6}(1)(2)(3) = 1}\)
\(\bullet\ \) ស្វ៊ែ \( OABC\colon \boxed{x^2+y^2+z^2-x-2y-3z=0}\)

ដំណោះស្រាយ

លំហាត់ ៧៤៖
\(\bullet\ (L)\colon \boxed{x=1+2t, \; y=2+2t, \; z=-2+t}\)
\(\bullet\ J = (L)\cap(P) \implies \boxed{J(-1, 0, -3)}\)
\(\bullet\ d = IJ = 3, \; 2\pi r = 8\pi \implies r = 4\)
\(R = \sqrt{3^2+4^2} = 5\)៖
\(\boxed{(S)\colon (x-1)^2+(y-2)^2+(z+2)^2 = 25}\)

លំហាត់ ៧៥៖
\(\bullet\ (L)\colon \boxed{x=-1+2t, \; y=3-t, \; z=2+t}\)
\(\bullet\ H = (L)\cap(P) \implies 6t-2=0 \implies \boxed{H\left(-\frac{1}{3}, \frac{8}{3}, \frac{7}{3}\right)}\)
\(\bullet\ A' = 2H - A = \boxed{\left(\frac{1}{3}, \frac{7}{3}, \frac{8}{3}\right)}\)

៤. ដំណោះស្រាយលម្អិតនៃលំហាត់ប្រឡងបាក់ឌុប (២០០២–២០២៦)

ដំណោះស្រាយ

ក សមីការប៉ារ៉ាម៉ែត្របន្ទាត់ \((D)\) កាត់តាម \(O(0,0,0)\) មានវ៉ិចទ័រប្រាប់ទិស \(\vec{u}(1,1,1)\)៖

គ សមីការប្លង់ \((ABC)\) កាត់តាម \(A(3,0,1)\) មានវ៉ិចទ័រណរម៉ាល់ \(\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC} = (11, 5, 9)\)៖
\(11(x-3) + 5(y-0) + 9(z-1) = 0\)
\(11x + 5y + 9z - (33 + 0 + 9) = 0\)
ដូចនេះ \(\boxed{(ABC)\colon 11x + 5y + 9z - 42 = 0}\)

ឃ ផ្ទៃក្រឡាត្រីកោណ \(ABC\)៖
\(S_{ABC} = \frac{1}{2}\|\overrightarrow{AB}\times\overrightarrow{AC}\| = \frac{1}{2}\sqrt{11^2+5^2+9^2}\)
ដូចនេះ \(\boxed{S_{ABC} = \frac{\sqrt{227}}{2}\text{ (ឯកតាផ្ទៃ)}}\)

ង ប្រសព្វរវាងបន្ទាត់ \((D)\) និងប្លង់ \((ABC)\)៖
ជំនួសសមីការ \((D)\) ចូលក្នុងសមីការប្លង់ \((ABC)\)៖
\(11(t) + 5(t) + 9(t) - 42 = 0 \iff 25t = 42 \implies t = \frac{42}{25}\)
ដូចនេះ \(\boxed{I\left(\frac{42}{25}, \frac{42}{25}, \frac{42}{25}\right)}\)

ដំណោះស្រាយ

ក សមីការប៉ារ៉ាម៉ែត្របន្ទាត់ \((D)\) កាត់តាម \(A(2,0,2)\) ស្របនឹង \(\overrightarrow{BC} = (0-1, 2-2, 3-0) = (-1, 0, 3)\)៖

គ គណនា \(\overrightarrow{AB}\times\overrightarrow{AC}\)៖
\(\overrightarrow{AB}\times\overrightarrow{AC} = (2-(-4), -(-1-4), -2-(-4)) = \boxed{(6, 5, 2)}\)
ដោយ \(\overrightarrow{AB}\times\overrightarrow{AC} \neq \vec{0} \implies A,B,C\) មិនរត់ត្រង់គ្នា។
ប្លង់ \((ABC)\) កាត់ \(A(2,0,2)\) មាន \(\vec{n} = (6,5,2)\)៖
\(6(x-2) + 5(y-0) + 2(z-2) = 0\)
ដូចនេះ \(\boxed{(ABC)\colon 6x + 5y + 2z - 16 = 0}\)
\(S_{ABC} = \frac{1}{2}\sqrt{6^2+5^2+2^2} = \boxed{\frac{\sqrt{65}}{2}\text{ cm}^2}\)

ឃ មាឌតេត្រាអែត \(OABC\)៖
\(V_{OABC} = \frac{1}{6}|[\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}]| = \frac{1}{6}|\overrightarrow{OA}\cdot(\overrightarrow{AB}\times\overrightarrow{AC})|\)
\(= \frac{1}{6}|2(6) + 0(5) + 2(2)| = \frac{16}{6} = \boxed{\frac{8}{3}\text{ cm}^3}\)
ចម្ងាយពីចំណុច \(O(0,0,0)\) ទៅប្លង់ \((ABC)\)៖
\(d(O, (ABC)) = \frac{|-16|}{\sqrt{65}} = \boxed{\frac{16\sqrt{65}}{65}\text{ cm}}\)

ដំណោះស្រាយ

ក សមីការប្លង់ \((P)\) កាត់តាម \(A(3,2,0)\) មានវ៉ិចទ័រណរម៉ាល់ \(\overrightarrow{AK} = (0-3, -1-2, 3-0) = (-3,-3,3) = -3(1,1,-1)\)៖
\(1(x-3) + 1(y-2) - 1(z-0) = 0\)
ដូចនេះ \(\boxed{(P)\colon x + y - z - 5 = 0}\)

ខ ផ្ទៀងផ្ទាត់ចំណុច \(B, C, D\) លើប្លង់ \((P)\)៖
\(\bullet\ B(5,0,0)\colon 5 + 0 - 0 - 5 = 0\) (ពិត) \(\implies B\in(P)\)
\(\bullet\ C(0,5,0)\colon 0 + 5 - 0 - 5 = 0\) (ពិត) \(\implies C\in(P)\)
\(\bullet\ D(0,0,-5)\colon 0 + 0 - (-5) - 5 = 0\) (ពិត) \(\implies D\in(P)\)
ដូចនេះ \(B, C\) និង \(D\) ជាចំណុចនៅលើប្លង់ \((P)\)។

គ គណនាផ្ទៃក្រឡាត្រីកោណ \(BCD\)៖
\(\overrightarrow{BC} = (-5, 5, 0), \; \overrightarrow{BD} = (-5, 0, -5)\)
\(\overrightarrow{BC}\times\overrightarrow{BD} = (-25-0, -(25-0), 0-(-25)) = (-25, -25, 25)\)
\(S_{BCD} = \frac{1}{2}\|\overrightarrow{BC}\times\overrightarrow{BD}\| = \frac{1}{2}\sqrt{25^2+25^2+25^2}\)
ដូចនេះ \(\boxed{S_{BCD} = \frac{25\sqrt{3}}{2}\text{ (ឯកតាផ្ទៃ)}}\)

ឃ ប្រវែង \(AK\) និងមាឌតេត្រាអែត \(KBCD\)៖
\(AK = \sqrt{(-3)^2+(-3)^2+3^2} = \sqrt{27} = \boxed{3\sqrt{3}}\)
ដោយ \(AK\perp (P)\) ហើយ \(B, C, D\in(P)\) នោះ \(AK\) ជាកម្ពស់នៃតេត្រាអែត \(KBCD\) ធៀបនឹងបាត \(BCD\)៖
\(V_{KBCD} = \frac{1}{3} S_{BCD} \cdot AK = \frac{1}{3} \left(\frac{25\sqrt{3}}{2}\right)(3\sqrt{3})\)
ដូចនេះ \(\boxed{V_{KBCD} = \frac{75}{2}\text{ (ឯកតាមាឌ)}}\)

ដំណោះស្រាយ

ក \(\bullet\) ជំនួស \(A(0,2,2)\) ចូល \((\alpha)\)៖
\(0 - 2 - 2 + 4 = 0\) (ពិត) \(\implies A\in (\alpha)\)
\(\bullet\) ប្លង់ \((\alpha)\) មានវ៉ិចទ័រណរម៉ាល់ \(\vec{n}_\alpha = (1, -1, -1)\)។
បន្ទាត់ \((D)\) មានវ៉ិចទ័រប្រាប់ទិស \(\vec{u}_D = (1, -1, -1)\)។
ដោយ \(\vec{n}_\alpha = \vec{u}_D \implies (\alpha)\perp (D)\)។

ខ រកកូអរដោនេចំណុច \(M, N, P\)៖
\(\bullet\ M = (\alpha)\cap Ox (y=0, z=0)\colon x + 4 = 0 \implies \boxed{M(-4, 0, 0)}\)
\(\bullet\ N = (\alpha)\cap Oy (x=0, z=0)\colon -y + 4 = 0 \implies \boxed{N(0, 4, 0)}\)
\(\bullet\ P = (\alpha)\cap Oz (x=0, y=0)\colon -z + 4 = 0 \implies \boxed{P(0, 0, 4)}\)

គ គណនាប្រវែងជ្រុងទាំងបីនៃ \(\triangle MNP\)៖
\(MN = \sqrt{(0-(-4))^2+(4-0)^2+0^2} = \sqrt{16+16} = 4\sqrt{2}\)
\(MP = \sqrt{(0-(-4))^2+0^2+(4-0)^2} = \sqrt{16+16} = 4\sqrt{2}\)
\(NP = \sqrt{0^2+(0-4)^2+(4-0)^2} = \sqrt{16+16} = 4\sqrt{2}\)
ដោយ \(MN = MP = NP = 4\sqrt{2} \implies \boxed{\triangle MNP \text{ ជាត្រីកោណសម័ង្ស}}\)

ឃ \(\overrightarrow{MP} = (4, 0, 4), \; \overrightarrow{MN} = (4, 4, 0)\)៖
\(\overrightarrow{MP}\cdot\overrightarrow{MN} = 4(4) + 0(4) + 4(0) = \boxed{16}\)
\(\cos\angle PMN = \frac{\overrightarrow{MP}\cdot\overrightarrow{MN}}{MP\cdot MN} = \frac{16}{(4\sqrt{2})^2} = \frac{1}{2} \implies \angle PMN = 60^\circ\)
\(S_{MNP} = \frac{\sqrt{3}}{4}(MN)^2 = \frac{\sqrt{3}}{4}(32) = \boxed{8\sqrt{3}\text{ (ឯកតាផ្ទៃ)}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{OB} = \overrightarrow{OA}+\overrightarrow{AB} = (1-1, 0+2, 1+1) = (0,2,2) \implies \boxed{B(0,2,2)}\)
ប្លង់ \((P)\) កាត់តាម \(A(1,0,1)\) មាន \(\vec{n} = \overrightarrow{AB} = (-1,2,1)\)៖
\(-1(x-1) + 2(y-0) + 1(z-1) = 0\)
ដូចនេះ \(\boxed{(P)\colon -x + 2y + z = 0}\) (ឬ \(x-2y-z=0\))

ខ \(\overrightarrow{AC} = (2-1, 1-0, 0-1) = \boxed{(1, 1, -1)}\)
\(\overrightarrow{CD} = (1-2, 3-1, 1-0) = \boxed{(-1, 2, 1)}\)
\(\overrightarrow{AB}\cdot\overrightarrow{AC} = -1(1) + 2(1) + 1(-1) = 0 \implies \overrightarrow{AB}\perp\overrightarrow{AC}\)
ដោយ \(\overrightarrow{CD} = \overrightarrow{AB} = (-1,2,1) \implies ABCD\) ជាប្រលេឡូក្រាម ដែលមានមុំ \(\angle A = 90^\circ \implies \boxed{ABCD \text{ ជាចតុកោណកែង}}\)

គ គណនា \(\overrightarrow{AB}\times\overrightarrow{AC}\) និងផ្ទៃ \(S_{ABCD}\)៖
\(\overrightarrow{AB}\times\overrightarrow{AC} = (-2-1, -(1-1), -1-2) = \boxed{(-3, 0, -3)}\)
\(S_{ABCD} = \|\overrightarrow{AB}\times\overrightarrow{AC}\| = \sqrt{(-3)^2+0^2+(-3)^2}\)
ដូចនេះ \(\boxed{S_{ABCD} = 3\sqrt{2}\text{ (ឯកតាផ្ទៃ)}}\)

ឃ មាឌចតុមុខ \(OABC\)៖
\(V_{OABC} = \frac{1}{6}|[\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}]| = \frac{1}{6}|\overrightarrow{OA}\cdot(\overrightarrow{AB}\times\overrightarrow{AC})|\)
\(= \frac{1}{6}|1(-3)+0+1(-3)| = \frac{6}{6} = \boxed{1\text{ (ឯកតាមាឌ)}}\)
ប្លង់ \((ABC)\) កាត់ \(A(1,0,1)\) មាន \(\vec{n} = (1,0,1)\colon x+z-2=0\)
\(d(O, (ABC)) = \frac{|0+0-2|}{\sqrt{1+0+1}} = \frac{2}{\sqrt{2}} = \boxed{\sqrt{2}}\)

ដំណោះស្រាយ

ក \((D)\colon \begin{cases} x = t \\ y = t \\ z = 3 + t \end{cases} \quad{} (t\in\mathbb{R})\)
ខ \(\overrightarrow{SM} = (x, y, z-3) \implies \vec{u}\cdot\overrightarrow{SM} = 1(x)+1(y)+1(z-3) = \boxed{x + y + z - 3}\)
\(\overrightarrow{SM}\perp\vec{u} \iff \vec{u}\cdot\overrightarrow{SM}=0 \iff \boxed{x + y + z - 3 = 0}\)
សំណុំចំណុច \(M\) គឺជា ប្លង់មួយកាត់តាមចំណុច \(S\) ហើយកែងនឹងវ៉ិចទ័រ \(\vec{u}\)។

គ \(SA = \sqrt{3^2+0+(-3)^2} = \sqrt{18} = 3\sqrt{2}\)
\(SB = \sqrt{0+3^2+(-3)^2} = \sqrt{18} = 3\sqrt{2}\)
\(AB = \sqrt{(-3)^2+3^2+0} = \sqrt{18} = 3\sqrt{2}\)
ដោយ \(SA = SB = AB = 3\sqrt{2} \implies \boxed{\triangle SAB \text{ ជាត្រីកោណសម័ង្ស}}\)

ឃ \(\overrightarrow{SA} = (3,0,-3), \; \overrightarrow{SB} = (0,3,-3)\)
\(\overrightarrow{SA}\times\overrightarrow{SB} = (0-(-9), -(-9-0), 9-0) = \boxed{(9, 9, 9)}\)
\(S_{SAB} = \frac{1}{2}\|\overrightarrow{SA}\times\overrightarrow{SB}\| = \frac{1}{2}\sqrt{81+81+81} = \boxed{\frac{9\sqrt{3}}{2}\text{ (ឯកតាផ្ទៃ)}}\)

ង \(OS = 3, OA = 3, OB = 3 \implies R = 3\)
សមីការស្វ៊ែផ្ចិត \(O(0,0,0)\) កាំ \(R=3\)៖ \(\boxed{x^2 + y^2 + z^2 = 9}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AM}=(x+1, y-1, z), \overrightarrow{BM}=(x-2, y, z-1)\)
\(\overrightarrow{AM}\cdot\overrightarrow{BM} = (x+1)(x-2) + y(y-1) + z(z-1) = 0\)
ដូចនេះ \(\boxed{x^2 + y^2 + z^2 - x - y - z - 2 = 0}\)
សំណុំចំណុច \(M\) គឺជា ស្វ៊ែដែលមានអង្កត់ផ្ចិត \([AB]\)។

ខ \(\overrightarrow{BC} = (-1, 2, 1), \; \overrightarrow{BD} = (-2, 1, -3)\)
\(\overrightarrow{BC}\times\overrightarrow{BD} = (-6-1, -(3+2), -1+4) = \boxed{(-7, -5, 3)}\)
\(S = \frac{1}{2}\|\overrightarrow{BC}\times\overrightarrow{BD}\| = \frac{\sqrt{49+25+9}}{2} = \frac{\sqrt{83}}{2}\)
\(S = \sqrt[3]{e^m} = e^{m/3} \implies \frac{m}{3} = \ln\left(\frac{\sqrt{83}}{2}\right) \implies \boxed{m = 3\ln\left(\frac{\sqrt{83}}{2}\right)}\)

គ ប្លង់ \((BCD)\) កាត់ \(B(2,0,1)\) មាន \(\vec{n} = (-7,-5,3)\)៖
\(-7(x-2) - 5(y-0) + 3(z-1) = 0\)
ដូចនេះ \(\boxed{(BCD)\colon 7x + 5y - 3z - 11 = 0}\)
សមីការបន្ទាត់៖ \(x = 1+t, y = -1-t, z = t\)
ជំនួសចូល៖ \(7(1+t) + 5(-1-t) - 3t - 11 = 0\)
\(-t - 9 = 0 \implies t = -9 \implies \boxed{N(-8, 8, -9)}\)

ឃ ចម្ងាយពី \(A(-1,1,0)\) ទៅប្លង់ \((BCD)\)៖
\(d(A,(BCD)) = \frac{|7(-1)+5(1)-3(0)-11|}{\sqrt{7^2+5^2+(-3)^2}} = \boxed{\frac{13}{\sqrt{83}}}\)
\(V_{ABCD} = \frac{1}{3} S_{BCD} \cdot d = \frac{1}{3}\cdot\frac{\sqrt{83}}{2}\cdot\frac{13}{\sqrt{83}} = \boxed{\frac{13}{6}\text{ (ឯកតាមាឌ)}}\)

ដំណោះស្រាយ

ក ផ្ចិត \(I = \) មធ្យម \( AB = (1,2,1)\), កាំ \(R = \frac{AB}{2} = \sqrt{2}\)៖
\((S)\colon \boxed{(x-1)^2 + (y-2)^2 + (z-1)^2 = 2}\)
ជំនួស \((d)\) ចូល៖ \(t^2 + 0^2 + t^2 = 2 \iff 2t^2 = 2 \implies t = \pm 1\)
ដូចនេះ ចំណុចប្រសព្វគឺ \(\boxed{(2,2,2)}\) និង \(\boxed{(0,2,0)}\)

ខ \(\overrightarrow{CA} = (1,2,-1), \; \overrightarrow{CB} = (-1,2,1)\)
\(\vec{n} = \overrightarrow{CA}\times\overrightarrow{CB} = (2-(-2), -(1-1), 2-(-2)) = \boxed{(4, 0, 4)}\)
\(S_{ABC} = \frac{1}{2}\|\vec{n}\| = \frac{1}{2}\sqrt{16+0+16} = \boxed{2\sqrt{2}\text{ (ឯកតាផ្ទៃ)}}\)
ប្លង់ \((ABC)\colon 4(x-1) + 0 + 4(z-1) = 0 \iff \boxed{x + z - 2 = 0}\)

គ \(\bullet\ M = (ABC)\cap Ox (y=0, z=0) \implies \boxed{M(2,0,0)}\)
\(\bullet\ N = (ABC)\cap Oz (x=0, y=0) \implies \boxed{N(0,0,2)}\)
\(\overrightarrow{AB} = (-2,0,2), \; \overrightarrow{MN} = (-2,0,2) \implies \overrightarrow{AB} = \overrightarrow{MN}\)
\(\implies ABNM\) ជាប្រលេឡូក្រាម។
\(\overrightarrow{AM} = (0,-2,0) \implies \overrightarrow{AB}\cdot\overrightarrow{AM} = 0 \implies \angle A = 90^\circ\)
ដូចនេះ \(\boxed{ABNM \text{ ជាចតុកោណកែង}}\)

ឃ ចម្ងាយពី \(D(0,2,0)\) ទៅប្លង់ \((ABC)\)៖
\(d(D,(ABC)) = \frac{|0+0-2|}{\sqrt{1^2+0+1^2}} = \frac{2}{\sqrt{2}} = \boxed{\sqrt{2}}\)
\(V_{DCBA} = \frac{1}{3} S_{ABC}\cdot d = \frac{1}{3}(2\sqrt{2})\sqrt{2} = \boxed{\frac{4}{3}\text{ (ឯកតាមាឌ)}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (2,-8,-2), \; \overrightarrow{AC} = (3,0,1)\)
\(\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC} = (-8, -(2+6), 24) = \boxed{(-8, -8, 24)}\)
\(\vec{n} = -8(1, 1, -3) \implies\) យក \((1,1,-3)\) ជាណរម៉ាល់នៃ \((P)\)៖
\(1(x+1) + 1(y-2) - 3(z-1) = 0 \iff \boxed{(P)\colon x + y - 3z + 2 = 0}\)
បន្ទាត់ \((d)\colon \begin{cases} x = -2+3t \\ y = 0 \\ z = t \end{cases} \quad{} (t\in\mathbb{R})\)

ខ \(x^2 + (y-1)^2 + (z+1)^2 = 2 + 1 + 1 = 4 = 2^2\)
\(\implies \) ផ្ចិត \( \boxed{I(0, 1, -1)}, \quad{} \) កាំ \( \boxed{r = 2}\)

ប្រសព្វរវាង \((S)\) និង \((d)\)៖
\((-2+3t)^2 + (0-1)^2 + (t+1)^2 = 4\)
\(4 - 12t + 9t^2 + 1 + t^2 + 2t + 1 = 4 \iff 5t^2 - 5t + 1 = 0\)
\(\Delta = 25 - 20 = 5 \implies t = \frac{5\pm\sqrt{5}}{10}\)
ដូចនេះ \(\boxed{\left(-2+\frac{3(5\pm\sqrt{5})}{10}, 0, \frac{5\pm\sqrt{5}}{10}\right)}\)

គ ផ្ទៃក្រឡាត្រីកោណ \(ABC\)៖
\(S_{ABC} = \frac{1}{2}\|\vec{n}\| = \frac{1}{2}\sqrt{64+64+576} = \frac{\sqrt{704}}{2} = \boxed{4\sqrt{11}}\)
ចម្ងាយពី \(D(0,1,-1)\) ទៅប្លង់ \((P)\)៖
\(d(D,(P)) = \frac{|0+1-3(-1)+2|}{\sqrt{1^2+1^2+(-3)^2}} = \frac{6}{\sqrt{11}} = \boxed{\frac{6\sqrt{11}}{11}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{BC} = (-4, 4, -5) \implies \|\overrightarrow{BC}\| = \sqrt{16+16+25} = \sqrt{57}\)
\(r = \frac{\sqrt{57}}{2} \implies r^2 = \frac{57}{4}\)៖
ដូចនេះ \(\boxed{(S)\colon (x-1)^2 + (y-2)^2 + (z-4)^2 = \frac{57}{4}}\)

ខ \(\overrightarrow{AC} = (-3, 1, -6)\)។ ប្លង់ \((P)\) កាត់តាម \(B(2,-1,3)\)៖
\(-3(x-2) + 1(y+1) - 6(z-3) = 0\)
ដូចនេះ \(\boxed{(P)\colon 3x - y + 6z - 25 = 0}\)

គ \(\overrightarrow{AB} = (1, -3, -1)\)។ បន្ទាត់ \((d)\) កាត់តាម \(C(-2,3,-2)\)៖

ឃ \(d(B,A) = \|\overrightarrow{AB}\| = \sqrt{1+9+1} = \boxed{\sqrt{11}}\)
\(\overrightarrow{CB} = (4, -4, 5) \implies \overrightarrow{CB}\times\overrightarrow{AB} = (19, 9, -8)\)
\(\|\overrightarrow{CB}\times\overrightarrow{AB}\| = \sqrt{361+81+64} = \sqrt{506}\)
\(d(B,(d)) = \frac{\|\overrightarrow{CB}\times\overrightarrow{AB}\|}{\|\overrightarrow{AB}\|} = \frac{\sqrt{506}}{\sqrt{11}} = \boxed{\sqrt{46}}\)

ង \(\overrightarrow{BA} = (-1, 3, 1), \; \overrightarrow{BC} = (-4, 4, -5)\)
\(\overrightarrow{BA}\times\overrightarrow{BC} = (-19, -9, 8)\)
\(S_{\text{ប្រលេ}} = \|\overrightarrow{BA}\times\overrightarrow{BC}\| = \boxed{\sqrt{506}\text{ (ឯកតាផ្ទៃ)}}\)
\(S_\triangle = \frac{1}{2}S_{\text{ប្រលេ}} = \boxed{\frac{\sqrt{506}}{2}\text{ (ឯកតាផ្ទៃ)}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{MN} = (1, 1, 1), \; \overrightarrow{MP} = (2, 2, -2)\)
\(\vec{n} = \overrightarrow{MN}\times\overrightarrow{MP} = (-2-2, -(-2-2), 2-2) = \boxed{(-4, 4, 0)}\)
យក \(\vec{n}_\beta = (1, -1, 0)\) ជាវ៉ិចទ័រណរម៉ាល់នៃ \((\beta)\)៖
\(1(x+1) - 1(y-0) + 0(z-1) = 0\)
ដូចនេះ \(\boxed{(\beta)\colon x - y + 1 = 0}\)

ខ កាំ \(R = MN = \sqrt{1^2+1^2+1^2} = \sqrt{3} \implies R^2 = 3\)
សមីការស្វ៊ែ \((S)\) ផ្ចិត \(M(-1,0,1)\)៖
ដូចនេះ \(\boxed{(S)\colon (x+1)^2 + y^2 + (z-1)^2 = 3}\)
ចម្ងាយពីផ្ចិត \(M\) ទៅប្លង់ \((\alpha)\)៖
\(d = \frac{|-1 - 2(0) + 1 - 4|}{\sqrt{1^2+(-2)^2+1^2}} = \frac{4}{\sqrt{6}} = \frac{2\sqrt{6}}{3}\)
ប្រៀបធៀប៖ \(d^2 = \frac{16}{6} = \frac{8}{3} < 3 = R^2 \implies d < R\)
ដូចនេះ ប្លង់ \((\alpha)\) កាត់ស្វ៊ែ \((S)\) បានជាខ្សែរង្វង់មួយ។

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (-1, 2, 1), \; \overrightarrow{AC} = (1, 1, -1)\)
\(\overrightarrow{AB}\cdot\overrightarrow{AC} = -1(1) + 2(1) + 1(-1) = 0\)
ដូចនេះ \(\boxed{\triangle ABC \text{ ជាត្រីកោណកែងត្រង់កំពូល } A}\)

ខ \(\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC} = (-2-1, -(1-1), -1-2) = \boxed{(-3, 0, -3)}\)
យក \(\vec{n}_{ABC} = (1, 0, 1)\) ជាណរម៉ាល់កាត់ \(A(1,0,1)\)៖
\(1(x-1) + 0(y-0) + 1(z-1) = 0 \iff \boxed{x + z - 2 = 0}\)

គ បន្ទាត់ \((d)\perp (ABC) \implies \vec{u}_d = (1, 0, 1)\)៖

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (2, -4, -1), \; \overrightarrow{DC} = (2, -4, -1)\)
\(\implies \overrightarrow{AB} = \overrightarrow{DC} \implies \boxed{ABCD \text{ ជាប្រលេឡូក្រាម}}\)
\(\overrightarrow{AD} = (-1, 3, 1) \implies \overrightarrow{AB}\times\overrightarrow{AD} = (-1, -1, 2)\)
\(S_{ABCD} = \|\overrightarrow{AB}\times\overrightarrow{AD}\| = \sqrt{1+1+4} = \boxed{\sqrt{6}}\)

ខ \((AB)\colon x = 2 + 2t, \; y = 2 - 4t, \; z = 1 - t \; (t\in\mathbb{R})\)
គ ប្លង់ \((ABD)\) កាត់ \(A(2,2,1)\) មាន \(\vec{n} = (-1,-1,2)\)៖
\(-(x-2) - (y-2) + 2(z-1) = 0 \iff \boxed{x + y - 2z - 2 = 0}\)

ដំណោះស្រាយ

ក \(\vec{u}+\vec{v} = (1-1, -1+2, 2+2) = \boxed{(0,1,4)}\)
ខ \(\vec{w}\times\vec{u} = \boxed{-4\vec{j}-2\vec{k} = (0,-4,-2)}\)

គ \(\vec{w}\times\vec{v} = \boxed{6\vec{i}+3\vec{k} = (6,0,3)}\)

ដំណោះស្រាយ

ក \(\overrightarrow{MN} = (3-2, 5-3, 6-4) = \boxed{(1, 2, 2)}\)
\(\overrightarrow{QP} = (4-3, 6-4, 7-5) = \boxed{(1, 2, 2)}\)

ខ ដោយ \(\overrightarrow{MN} = \overrightarrow{QP} = (1, 2, 2)\)
\(\implies \boxed{MNPQ \text{ ជាប្រលេឡូក្រាម}}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (2, 0, 1), \; \overrightarrow{AC} = (3, 1, 2)\)
\(\overrightarrow{AD} = (2, 2, -4), \; \overrightarrow{BC} = (1, 1, 1)\)
\(\overrightarrow{BD} = (0, 2, -5), \; \overrightarrow{CD} = (-1, 1, -6)\)
ខ \(AB = \sqrt{4+0+1} = \boxed{\sqrt{5}}\)
\(AC = \sqrt{9+1+4} = \boxed{\sqrt{14}}\)
\(AD = \sqrt{4+4+16} = \boxed{2\sqrt{6}}\)
\(BC = \sqrt{1+1+1} = \boxed{\sqrt{3}}\)
\(BD = \sqrt{0+4+25} = \boxed{\sqrt{29}}\)
\(CD = \sqrt{1+1+36} = \boxed{\sqrt{38}}\)

\(\bullet\ AB^2 + AD^2 = 5 + 24 = 29 = BD^2\)
\(\implies \boxed{\triangle ABD \text{ ជាត្រីកោណកែងត្រង់ } A}\)
\(\bullet\ AC^2 + AD^2 = 14 + 24 = 38 = CD^2\)
\(\implies \boxed{\triangle ACD \text{ ជាត្រីកោណកែងត្រង់ } A}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB} = (2, -2, -2)\)
\(\overrightarrow{AC} = (-2, -2, -2)\)
\(\overrightarrow{AD} = (1, -1, -1)\)
\(\overrightarrow{BC} = (-4, 0, 0)\)

ខ \(\vec{n}\cdot\overrightarrow{AB} = 0(2)+1(-2)-1(-2) = \boxed{0}\)
\(\vec{n}\cdot\overrightarrow{AC} = 0(-2)+1(-2)-1(-2) = \boxed{0}\)
ដោយ \(\vec{n}\neq\vec{0}\) កែងនឹងវ៉ិចទ័រពីរ \(\overrightarrow{AB}, \overrightarrow{AC}\) មិនកូលីនេអ៊ែរ
\(\implies \boxed{\vec{n}=(0,1,-1) \text{ ជាវ៉ិចទ័រណរម៉ាល់នៃប្លង់ } (ABC)}\)

ដំណោះស្រាយ

ផ្នែក A៖
ក \(\overrightarrow{AB}=(6,0,-6), \; \overrightarrow{AC}=(0,6,-6), \; \overrightarrow{BC}=(-6,6,0)\)
\(\overrightarrow{AB}\times\overrightarrow{AC} = (36,36,36)\neq\vec{0}\)
\(\implies A,B,C \) កំណត់បានប្លង់មួយ
\(\vec{n}\cdot\overrightarrow{AB}=6-6=0, \; \vec{n}\cdot\overrightarrow{AC}=6-6=0\)
\(\implies \vec{n}(1,1,1) \) ជាវ៉ិចទ័រណរម៉ាល់នៃ \( (ABC)\)។
\((ABC)\colon 1(x-1)+1(y+1)+1(z-4)=0\)
\(\iff \boxed{x + y + z - 4 = 0}\)
ខ \(\|\overrightarrow{AB}\|=\|\overrightarrow{AC}\|=\|\overrightarrow{BC}\|=6\sqrt{2}\)
\(\implies \boxed{\triangle ABC \text{ ជាត្រីកោណសម័ង្ស}}\)

ផ្នែក B៖
\((y+2)^2-4-8x-12=0 \iff \boxed{(y+2)^2 = 8(x+2)}\)
ទម្រង់ \((y-k)^2=4p(x-h) \implies p=2\)
\(\bullet\) កំពូល \(\boxed{V(-2,-2)}\) \(\bullet\) កំណុំ \(\boxed{F(0,-2)}\)
\(\bullet\) បន្ទាត់ប្រាប់ទិស \(\boxed{x = -4}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}=(-4,3,2), \; \overrightarrow{BC}=(4,2,5)\)
\(\overrightarrow{CD}=(4,-3,-2), \; \overrightarrow{AD}=(4,2,5)\)
\(\bullet\ \overrightarrow{AB}=\overrightarrow{DC} \implies ABCD\) ជាប្រលេឡូក្រាម
\(\bullet\ \overrightarrow{AB}\cdot\overrightarrow{BC} = -16+6+10 = 0 \implies \angle B = 90^\circ\)
\(\bullet\ AB = \sqrt{16+9+4} = \sqrt{29}\)
\(\ BC = \sqrt{16+4+25} = 3\sqrt{5} \neq AB\)
\(\implies \boxed{ABCD \text{ ចតុកោណកែង មិនមែនការេ}}\)

ខ \(9y^2=25(9-x^2) \iff 25x^2+9y^2=225\)
\(\iff \boxed{\frac{x^2}{9} + \frac{y^2}{25} = 1} \) (អេលីបឈរ ផ្ចិត \( O)\)
\(a=5, b=3, c=\sqrt{25-9}=4\)៖
\(\bullet\) អ័ក្សធំ \(2a = \boxed{10}\), អ័ក្សតូច \(2b = \boxed{6}\)
\(\bullet\) កំពូល \(\boxed{V(0,\pm 5)}\) \(\bullet\) កំណុំ \(\boxed{F(0,\pm 4)}\)

ដំណោះស្រាយ

ក 1. \(\overrightarrow{MN}=(1,1,1) \implies MN=\boxed{\sqrt{3}}\)
\(\overrightarrow{NP}=(-1,1,-6) \implies NP=\boxed{\sqrt{38}}\)
\(\overrightarrow{NQ}=(-3,-1,-2) \implies NQ=\boxed{\sqrt{14}}\)
\(\overrightarrow{PQ}=(-2,-2,4) \implies PQ=\boxed{2\sqrt{6}}\)
2. \(\overrightarrow{NQ}\cdot\overrightarrow{PQ} = 6+2-8 = \boxed{0} \implies \boxed{\triangle NPQ \text{ កែងត្រង់ } Q}\)
\(S_{NPQ} = \frac{1}{2}NQ\cdot PQ = \frac{1}{2}\sqrt{14}\cdot 2\sqrt{6} = \boxed{2\sqrt{21}}\)

ខ \((y+2)^2 = -8(x-2) \implies (y-k)^2=4p(x-h)\)
\(h=2, k=-2, 4p=-8 \implies p=-2 < 0\)
\(\bullet\) កំពូល \(\boxed{V(2,-2)}\) \(\bullet\) កំណុំ \(\boxed{F(0,-2)}\)
\(\bullet\) បន្ទាត់ប្រាប់ទិស \(\boxed{x = 4}\)

ដំណោះស្រាយ

ក 1. \(\overrightarrow{AB}=(-6,6,-2), \overrightarrow{AC}=(-5,6,0)\)
\(\overrightarrow{AB}\times\overrightarrow{AC} = (12,10,-6)\neq\vec{0} \implies A,B,C \) មិនរត់ត្រង់គ្នា
2. \(\overrightarrow{BC}=(1,0,2), \overrightarrow{CD}=(6,5,-3)\)
\(\overrightarrow{BC}\cdot\overrightarrow{CD} = 6+0-6 = 0 \implies \boxed{\triangle BCD \text{ កែងត្រង់ } C}\)
\(BC = \sqrt{5}, CD = \sqrt{70}\)
\(S_{BCD} = \frac{1}{2}\sqrt{5}\sqrt{70} = \boxed{\frac{5\sqrt{14}}{2}\text{ (ឯកតាផ្ទៃ)}}\)

ខ \(12xy+36-(9x^2+12xy+4y^2)=0\)
\(\iff 9x^2+4y^2=36 \iff \boxed{\frac{x^2}{4}+\frac{y^2}{9}=1}\)
អេលីបឈរ \(a=3, b=2\)៖
\(\bullet\) អ័ក្សធំ \(2a=\boxed{6}\), អ័ក្សតូច \(2b=\boxed{4}\)
\(\bullet\) កំពូល \(\boxed{V(0,\pm 3)}\)

ដំណោះស្រាយ

ក \(\overrightarrow{AB}=(-3,\sqrt{3},0), \overrightarrow{AC}=(-3,-\sqrt{3},0)\)
\(\overrightarrow{BC}=(0,-2\sqrt{3},0)\)
\(AB = AC = BC = \sqrt{9+3} = 2\sqrt{3}\)
\(\implies \boxed{\triangle ABC \text{ ជាត្រីកោណសម័ង្ស}}\)
\(S_{ABC} = \frac{\sqrt{3}}{4}(2\sqrt{3})^2 = \boxed{3\sqrt{3}\text{ (ឯកតាផ្ទៃ)}}\)

ខ \(4x^2-12xy+9y^2=18y^2-12xy-36\)
\(\iff -4x^2+9y^2=36 \iff \boxed{\frac{y^2}{4}-\frac{x^2}{9}=1}\)
អ៊ីពែរបូលឈរ \(a=2, b=3, c=\sqrt{13}\)៖
\(\bullet\) កំពូល \(\boxed{V(0,\pm 2)}\) \(\bullet\) កំណុំ \(\boxed{F(0,\pm\sqrt{13})}\)
\(\bullet\) អាស៊ីមតូត \(\boxed{y = \pm\frac{2}{3}x}\)