c="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/5fdfea5b68eb.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0px auto;max-height:40vh" alt="JEE Main 2021 (Online) 26th August Morning Shift Mathematics - 3D Geometry Question 242 English Explanation">
$$ \begin{aligned} & \frac{x-1}{1}=\frac{y-3}{-2}=\frac{z-4}{-1}=\lambda_1 \\\\ & \left(\lambda_1+1\right)-2\left(-2 \lambda_1+3\right)-\left(-\lambda_1+4\right)=3 \\\\ & \Rightarrow 6 \lambda_1=12 \Rightarrow \lambda_1=2 \end{aligned} $$

So, foot of perpendicular from $(1,3,4)$ to plane $x-2 y-z=3$ is $A$ $(3,-1,2)$.

Let us also find the intersection point of the plane and line

$$ \begin{gathered} \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}=\lambda_2 \\\\ \left(2 \lambda_2+1\right)-2\left(\lambda_2+3\right)-\left(2 \lambda_2+4\right)=3-2 \lambda_2=12 \Rightarrow \lambda_2=-6 \end{gathered} $$

The intersection point of the plane and line is $B(-11,-3,-8)$ Line passing through $A$ and $B$ is

$$ \begin{aligned} & \frac{x-3}{-14}=\frac{y+1}{-2}=\frac{z-2}{-10}=\mu \\\\ & \frac{x-3}{7}=\frac{y+1}{1}=\frac{z-2}{5}=\mu \end{aligned} $$

Now, let's find the distance from $O(0,0,6)$ to this line $L$.

Let's say $C(7 \mu+3, \mu-1,5 \mu+2)$ is any point on $L$. Then,

$$ \begin{aligned} & \{(7 \mu+3)-0\} \cdot 7+\{(\mu-1)-0\} \cdot 1+\{(5 \mu+2)-6\} \cdot 5=0 \\\\ & \Rightarrow 49 \mu+21+\mu-1+25 \mu-20=0 \Rightarrow \mu=0 \\\\ & \therefore C(3,-1,2) \\\\ & \text { Distance }=\sqrt{(3-0)^2+(-1-0)^2+(2-6)^2}=\sqrt{26} \\\\ & d^2=26 \end{aligned} $$

3d Geometry

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