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$$\begin{aligned} & P(10,-2,-1) \\ & M N: \frac{x-2}{8}=\frac{y+5}{-12}=\frac{z-11}{16} \end{aligned}$$

General point

$$(8k + 2, - 12k - 5,16k + 11)$$

$$\overrightarrow {RQ} = (8k + 2 - 1)\widehat i + ( - 12k - 5 - 7)\widehat j + (16k + 11 - 6)\widehat k$$

$$\overrightarrow {RQ} = (8k + 1)\widehat i - (12k + 12)\widehat j + (16k + 5)\widehat k$$

$$\overrightarrow {RQ} \,.\,\overrightarrow {MN} = 0$$ (as both are perpendicular)

$$8(8 k+1)+12(12 k+12)+16(16 k+5)=0$$

$$\begin{aligned} & 64 k+8+144 k+144+256 k+80=0 \\ & 464 k=-232 \\ & k=\frac{-232}{464}=\frac{-1}{2} \\ & Q(-4+2,6-5,-8+11) \\ & Q(-2,1,3) \\ & P Q=\sqrt{(10+2)^2+(-3)^2+(4)^2} \\ & P Q=\sqrt{12^2+3^2+4^2} \\ & P Q=\sqrt{169}=13 \end{aligned}$$

19:T542,

JEE Main 2024 (Online) 6th April Morning Shift Mathematics - 3D Geometry Question 64 Hindi Explanation

$$\begin{aligned} & P(10,-2,-1) \\ & M N: \frac{x-2}{8}=\frac{y+5}{-12}=\frac{z-11}{16} \end{aligned}$$

व्यापक बिंदु

$$(8k + 2, - 12k - 5,16k + 11)$$

$$\overrightarrow {RQ} = (8k + 2 - 1)\widehat i + ( - 12k - 5 - 7)\widehat j + (16k + 11 - 6)\widehat k$$

$$\overrightarrow {RQ} = (8k + 1)\widehat i - (12k + 12)\widehat j + (16k + 5)\widehat k$$

$$\overrightarrow {RQ} \,.\,\overrightarrow {MN} = 0$$ (क्योंकि दोनों ल

3d Geometry

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