Two lines $L_1 : x = 5, \frac{y}{3- \alpha }=\frac{z}{-2}$ and $ L_2 : x = \alpha , \frac{y}{-1}=\frac{z}{2- \alpha }$ are coplanar. Then $\alpha $ can take value(s)

1, 4, 5

The equations of given lines are:

$L_1 : \frac{x-5}{0}=\frac{y}{3-\alpha}=\frac{z}{-2}, L_2 :\frac{x-\alpha}{0}=\frac{y}{-1}=\frac{z}{2-\alpha}$

These lines are coplanar. Therefore,

$\begin{vmatrix}5-\alpha  & 0-0 & 0-0\\0 & 3-\alpha & -2\\0 & -1 & 2-\alpha\end{vmatrix}=0$

$⇒ (5 - \alpha ) \begin{Bmatrix} (3- \alpha) (2 - \alpha ) -2\end{Bmatrix} = 0 $

$⇒ ((5 - \alpha ) (6 - 5\alpha + \alpha^2 - 2) = 0 $

$⇒ (5 - \alpha ) (\alpha^2 - 5\alpha + 4) = 0 $

$⇒( \alpha -1)( \alpha - 4) (\alpha - 5) = 0 ⇒ \alpha = 1, 4, 5$