Question

The line which contains all points $(x, y, z)$ which are of the form $(x, y, z)=(2,-2,5)+\lambda(1,-3,2)$ intersects the plane $2 x-3 y+4 z=163$ at $P$ and intersects the $Y Z$ -plane at $Q$. If the distance $P Q$ is $a \sqrt{b}$, where $a, b \in N$ and $a > 3$, then $(a < b)$ is equal to

• A. 23
• B. 95
• C. 27
• D. None of these

Answer 1

Answer: (A)

Equation of the line is $\frac{x-2}{1}=\frac{y+2}{-3}=\frac{z-5}{2}=\lambda$....(1)
Hence, any point on the line (1) can be taken as $x=\lambda+2$
$y=-(3 \lambda+2)$
$z=(2 \lambda+5)$
From some $\lambda$ point lies on the plane
$2 x-3 y+4 z=163$
$2(\lambda+2)+3(3 \lambda+2)+4(2 \lambda+5)=163$
$\Rightarrow 19 \lambda=133$
$ \Rightarrow \lambda=7$
Hence, $P \equiv(9,-23,19)$
Also, Eq. (1) intersect $Y Z$ -plane, i.e., $x=0$
$\lambda+2=0$
Hence, $\lambda=-2$
$\therefore Q(0,4,1)$
So, $P Q=\sqrt{9^{2}+27^{2}+18^{2}}$
$=9 \sqrt{1+3^{2}+2^{2}}=9 \sqrt{14}$
$\Rightarrow a=9$ and $b=14$
Hence, $a+b=9+14=23$.