Two lines L1:x=5,y3−α=z−2 and L2:x=α,y−1=z2−α are coplanar. Then α can take value(s)

$L_{1} : \frac{x - 5}{0} = \frac{y}{3 - α} = \frac{z}{- 2}$

$L_{2} : \frac{x - a}{0} = \frac{y}{- 1} = \frac{z}{2 - α}$

As $L_{1}, L_{2}$ are coplanar, therefore $|\begin{matrix} 5 - α & 0 & 0 \\ 0 & 3 - α & - 2 \\ 0 & - 1 & 2 - α \end{matrix}| = 0$

$\Rightarrow (5 - α) [6 - 5 α + α^{2} - 2] = 0$

$\Rightarrow (5 - α) (α - 1) (α - 4) = 0$

$\Rightarrow α = 1, 4, 5$

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