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 $2x-3y+4z=163$ at $P$ and intersects the $YZ$ -plane at $Q$. If the distance $PQ$ 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
$2x-3y+4z=163$
$2(\lambda +2)+3(3\lambda +2)+4(2\lambda +5)=163$
$\Rightarrow 19\lambda =133$
$\Rightarrow \lambda =7$
Hence, $P\equiv (9,-23,19)$
Also, Eq. (1) intersect $YZ$ -plane, i.e., $x=0$
$\lambda +2=0$
Hence, $\lambda =-2$
$\therefore Q(0,4,1)$
So, $PQ=\sqrt{9^2+27^2+18^2}$
$=9\sqrt{1+3^2+2^2}=9\sqrt{14}$
$\Rightarrow a=9$ and $b=14$
Hence, $a+b=9+14=23$.