intersects line (i)

$\therefore$ Direction ratio of $A B$ are

$$ \begin{array}{ll} & 0-2 \lambda-1,1-3 \lambda-2,2-\lambda-3 \\\\ &\text { i.e., } -2 \lambda-1,-3 \lambda-1,-\lambda-1 \end{array} $$

Line $L$ parallel to the plane $2 x+y-3 z=4$

$$ \begin{aligned} & \therefore 2(-2 \lambda-1)+1(-3 \lambda-1)-3(-\lambda-1)=0 \\\\ & \Rightarrow-4 \lambda-2-3 \lambda-1+3 \lambda+3=0 \\\\ & \Rightarrow -4 \lambda=0 \\\\ & \Rightarrow \lambda=0 \\\\ & \therefore \text { Direction ratio of } A B=-1,-1,-1 \\\\ & \text { i.e., } 1,1,1 \end{aligned} $$

$\therefore$ Equation of line $L$ is $\frac{x-0}{1}=\frac{y-1}{1}=\frac{z-2}{1}=\mu$

$$ \Rightarrow x=\mu, y=\mu+1, z=\mu+2 $$

JEE Main 2023 (Online) 6th April Evening Shift Mathematics - 3D Geometry Question 108 English Explanation

JEE Main 2023 (Online) 6th April Evening Shift Mathematics - 3D Geometry Question 108 English Explanation

$\therefore$ Direction ratios of $P Q=\mu-1, \mu+10, \mu$

Since, $P Q \perp L$

$$ \begin{aligned} & \therefore 1(\mu-1)+1(\mu+10)+\mu(1)=0 \\\\ & \Rightarrow \mu-1+\mu+10+\mu=0 \\\\ & \Rightarrow 3 \mu+9=0 \\\\ & \Rightarrow \mu=-3 \\\\ & \therefore Q \text { is }(-3,-2,-1) \end{aligned} $$

$\therefore$ Distance of the point $P(1,-9,2)$ from the line $L$ is

$$ \begin{array}{r} = \sqrt{(-3-1)^2+(-2+9)^2+(-1-2)^2} \\\\ = \sqrt{16+49+9}=\sqrt{74} \end{array} $$

3d Geometry

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