Let a line L pass through the point P ( 2, 3, 1 ) and be parallel...

Given,

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

Now direction ratio is given by,

\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 3 & - 2 \\ 1 & - 1 & 2\end{cases}\right| = 4\overset{^}{i} - 4\overset{^}{j} - 4\overset{^}{k}$</p><p>So, Equation of line will be$\dfrac{x - 2}{4} = \dfrac{y - 3}{- 4} = \dfrac{z - 1}{- 4}$</p><p>$⇒ \dfrac{x - 2}{1} = \dfrac{y - 3}{- 1} = \dfrac{z - 1}{- 1} = k$</p><p>Let$Q$be$( 5, 3, 8 ) and foot of $⊥$ from $Q$ on this line be $R$

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

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

Now using the perpendicular condition we get,

$( 1 ) ( k - 3 ) + ( - 1 ) ( - k ) + ( - 1 ) ( - k - 7 ) = 0$

$⇒ k = - \dfrac{4}{3}$

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

$⇒ 3{\alpha}^{2} = 158$