QJEEAdv12Ma1f08: JEE Advanced 2012 Mathematics - Class 12 JEEAdv12Ma1f08 Solution: The point is the intersection of the straight line joining the…

Equation of line QR is

$\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda$

General point on QR will be $(\lambda+2,4 \lambda+3, \lambda+5)$

Let $P \equiv(\lambda+2,4 \lambda+3, \lambda+5)$

Since, point P also lies on the plane.

$\begin{array}{rrrl}\therefore & 5(\lambda+2)-4(4 \lambda+3)-(\lambda+5) & =1 \\\Rightarrow & 5 \lambda+10-16 \lambda-12-\lambda-5 & =1 \\\Rightarrow & -12 \lambda-7 & =1\end{array}$

$\begin{array}{rrrl}\Rightarrow & \lambda &=\frac{-8}{12} \\\Rightarrow & \lambda &=\frac{-2}{3}\end{array}$

Hence,

$\begin{aligned}P & \equiv\left(\frac{-2}{3}+2,4\left(\frac{-2}{3}\right)+3, \frac{-2}{3}+5\right) \\& \equiv\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)\end{aligned}$

Now, for TS to be perpendicular to QR,

$\begin{array}{rlrl} & \lambda+4(4 \lambda+2)+(\lambda+1) & =0 \\\Rightarrow & \lambda \lambda+16 \lambda+8+\lambda+1 & =0 \\\Rightarrow & 18 \lambda =-9 \\\Rightarrow & \lambda =\frac{-1}{2}\end{array}$

$\begin{aligned}\therefore \quad \text { Point } S & \equiv\left(\frac{-1}{2}+2,4\left(\frac{-1}{2}\right)+3, \frac{-1}{2}+5\right) \\& \equiv\left(\frac{3}{2}, 1, \frac{9}{2}\right)\end{aligned}$

$\therefore$ Length of line segment PS

$\begin{aligned}& =\sqrt{\left(\frac{4}{3}-\frac{3}{2}\right)^2+\left(\frac{1}{3}-1\right)^2+\left(\frac{13}{3}-\frac{9}{2}\right)^2} \\& =\sqrt{\frac{1}{36}+\frac{4}{9}+\frac{1}{36}} \\& =\sqrt{\frac{1+16+1}{36}} \\& =\sqrt{\frac{18}{36}} \\& =\frac{1}{\sqrt{2}}\end{aligned}$