Question

If $L_1$ is the line of intersection of the planes $2x-2y+3z-2=0,x-y+z+1=0$ and $L_2$ is the line of intersection of the planes $x+2y-z-3=0,3x-y+1z-1=0,$ then the distance of the origin from the plane containing the lines $L_1$ and $L_2$ is:

• A. $2$
• B. $\frac{1}{3\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{5\sqrt{2}}$

Answer 1

The correct option is B $\frac{1}{3\sqrt{2}}$

Plane passing through line of intersection of first two planes is $(2x-2y+3z-2)+\lambda (x-y+z+1)=0$
$x(\lambda +2)-y(2+\lambda )+z(\lambda +3)+(\lambda -2)=0\cdots \left(i\right)$ is having infinite number of solutions with $x+2y-z-3=0$ and $3x-y+2z-1=0$ then
$\left|\begin{array}{ccc}\lambda +2& -(\lambda +2)& \lambda +3\\ 1& 2& -1\\ 3& -1& 2\end{array}\right|=0$
On solving, we have $\lambda =5$
Hence required equation of plane is $7x-7y+8z+3=0$
Now, perpendicular distance from $(0,0,0)$ is $\frac{3}{\sqrt{162}}=\frac{1}{3\sqrt{2}}$