Question

Let the line $L$ be the projection of the line $\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$ in the plane $x-2y-z=3.$ If $d$ is the distance of the point $(0,0,6)$ from $L,$ then $d^2$ is equal to

Answer 1

$L:\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$
$P_1:x-2y-z=3$
Equation of a plane $P_2$ which contains $L$ and perpendicular to $P_1,$
$\left|\begin{array}{ccc}x-1& y-3& z-4\\ 1& -2& -1\\ 2& 1& 2\end{array}\right|=0$
$\Rightarrow 3x+4y-5z+5=0$
Distances of point $(0,0,6)$ from $P_1$ and $P_2$ are
$d_1=\frac{25}{\sqrt{50}}=\sqrt{\frac{25}{2}}\text{and}{d}_2=\frac{9}{\sqrt{6}}=\sqrt{\frac{27}{2}}$
$\therefore d^2=d_1^2+d_2^2=26$