Question

Let $P$ be the plane containing the line $L_1:y+z=2,$ $x=0$ and is parallel to the line $L_2:x-z=2,y=0.$ If the distance of the plane $P$ from the origin is $d$ units, then the value of $3d^2$ is

Answer 1

Equation of $L_1$ is
$\frac{x-0}{0}=\frac{y-1}{1}=\frac{z-1}{-1}$
Similarly, equation of $L_2$ is $\frac{x-1}{1}=\frac{y-0}{0}=\frac{z+1}{1}$

So, equation of any plane through line $L_1$ is $ax+b(y-1)+c(z-1)=0\cdots \left(1\right)$
where $a\times 0+b\times 1+c\times (-1)=0\cdots \left(2\right)$
Also, the plane $\left(1\right)$ is parallel to the line $L_2.$
So $a\times 1+b\times 0+c\times 1=0\cdots \left(3\right)$

From $\left(2\right)$ and $\left(3\right),$ we get
$\frac{a}{1}=\frac{b}{-1}=\frac{c}{-1}$

So, the required plane $P,$ is
$1(x-0)-1(y-1)-1(z-1)=0$
$\Rightarrow x-y-z+2=0$

Distance from the origin to the plane $x-y-z+2=0$ is
$d=\left|\frac{2}{\sqrt{1+1+1}}\right|=\frac{2}{\sqrt{3}}$
Hence, $3d^2=4$