Question

Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$. If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^2$ is equal to_______

Answer 1

$\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 3 & -2 \\ 1 & -1 & 2\end{vmatrix}=4 \hat{ i }-4 \hat{ j }-4 \hat{ k }$
$\therefore$ Equation of line is $\frac{ x -2}{1}=\frac{ y -3}{-1}=\frac{ z -1}{-1}$
Let $Q$ be $(5,3,8)$ and foot of $\perp$ from $Q$ on this line be $R$.
Now, $R \equiv( k +2,- k +3,- k +1)$
$DR$ of $QR$ are $( k -3,- k ,- k -7)$
$ \therefore(1)( k -3)+(-1)(- k )+(-1)(- k -7)=0 $
$ \Rightarrow k =-\frac{4}{3}$
$\therefore \alpha^2=\left(\frac{13}{3}\right)^2+\left(\frac{4}{3}\right)^2+\left(\frac{17}{3}\right)^2=\frac{474}{9}$
$ \therefore 3 \alpha^2=158$