Given,

$x+3y-2z-2=0=x-y+2z$

Now direction ratio is given by,

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{\mathit{k}}\\ 1& 3& -2\\ 1& -1& 2\end{array}\right|=4\hat{\mathit{i}}-4\hat{\mathit{j}}-4\hat{\mathit{k}}$

So, Equation of line will be $\frac{x-2}{4}=\frac{y-3}{-4}=\frac{z-1}{-4}$

$\Rightarrow \frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-1}{-1}=k$

Let $Q$ be $(5,3,8)$ and foot of $\perp$ from $Q$ on this line be $R$

Now, $R\equiv (k+2,-k+3,-k+1)$

So, direction ratio of  $QR$ will be ($k-3,-k,-k-7$),

Now using the perpendicular condition we get,

$\left(1\right)(k-3)+(-1)(-k)+(-1)(-k-7)=0$

$\Rightarrow k=-\frac{4}{3}$

So, ${\alpha }^2={\left(\frac{13}{3}\right)}^2+{\left(\frac{4}{3}\right)}^2+{\left(\frac{17}{3}\right)}^2=\frac{474}{9}$

$\Rightarrow 3{\alpha }^2=158$