Question

The line which contains all the 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 $P$ intersects the $YZ$ plane at $Q$, If the distance $PQ$ is $a\sqrt{b}$ where $a,b$ $\in$ $N$ and $a>3$ then $(a+b)$ equals

• A. $23$
• B. $95$
• C. $27$
• D. $none$

Answer 1

The correct option is A $23$

$L:\frac{x-2}{1}=\frac{y+2}{-3}=\frac{z-5}{2}$

Plane$:2x-3y+4z=163$

say $L$ intersect plane at $(x,y,z)=(r+2),(-3r-2),(2r+5)$

$\therefore$ It must satisfy the equation of a plane.

$2(r+2)+3(3r+2)+4(2r+5)=163$

$\Rightarrow r=7$

$P(x,y,z)=(9,-23,19)$

$Q(x,y,z)$ is point of intersection of $L$ with $yz$ plane i.e.,$x=0$

$\therefore r+2=0r=-2$

$\therefore Q(x,y,z)=(0,4,1)$

$\therefore PQ=\sqrt{{(0-9)}^2+{(4+23)}^2+{(1-19)}^2}=9\sqrt{14}units=a\sqrt{b}$
$\therefore a=9,b=14\Rightarrow a+b=23$