Question

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

Answer 1

Answer: 158

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& -2\\ 1& -1& 2\end{array}\right|=4\hat{i}-4\hat{j}-4\hat{k}$
$\therefore$ 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 $(k-3,-k,-k-7)$
$\therefore (1)(k-3)+(-1)(-k)+(-1)(-k-7)=0$
$\Rightarrow k=-\frac{4}{3}$
$\therefore {\alpha }^2={\left(\frac{13}{3}\right)}^2+{\left(\frac{4}{3}\right)}^2+{\left(\frac{17}{3}\right)}^2=\frac{474}{9}$
$\therefore 3{\alpha }^2=158$