Question

Equation of lines
$L_1:2x-2y+3z-2=0=x-y+z+1=0$
$L_2:x+2y-z-3=0=3x-y+2z-1=0$
Distance of point $P(0,0,0)$ from the plane containing $L_1$ and $L_2$ and measured along the line $x=y=z$ is $\frac{a\sqrt{a}}{b}$ (where $a$ and $b$ are coprime numbers) then $a+b$ is

Answer 1

Equation of plane containing $L_1$ is -

$P_1+\lambda P_2=0\dots \dots \left(1\right)$

Equation of plane containing $L_2$ is -

$P_3+\mu P_4=0\dots \dots \left(2\right)$

For some $\lambda$ and $\mu$ equation (1) and (2) are respectively the same plane.

$x(2+\lambda )-(2+\lambda )y+(3+\lambda )z-2+\lambda =0$

$x(1+3\mu )+(2-\mu )y+(-1+2\mu )z-3-\mu =0$

$\frac{2+\lambda }{1+3\mu }=\frac{-(2+\lambda )}{2-\mu }=\frac{3+\lambda }{2\mu -1}=\frac{\lambda -2}{-(\mu +3)}$

$2-\mu =-1-3\mu$

$2\mu =-3\Rightarrow \mu =\frac{-3}{2}$

$x(1-\frac{9}{2})+(2+\frac{3}{2})y+(-1-3)z-3+\frac{3}{2}=0$

$-7x+7y-8z-3=0$

Let Q is $(\lambda ,\lambda ,\lambda )$ be a point in the plane

$\Rightarrow \lambda =\frac{-3}{8}$

So distance $=\sqrt{3}|\lambda |=\frac{3\sqrt{3}}{8}$