Two lines L_ 1 : x=5, (y)/(3-α)=(z)/(-2)L_ 2 : x=α, y -1 = z 2-α...

The equations of given lines can be written as $$\begin{array}{l}\mathrm{L}_{1}: \mathrm{x}-5=\frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}\\ \mathrm{ L}_{2}: \mathrm{x}-\alpha=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha} \end{array}$$ Since, these lines are coplanar. \begin{array}{l}\text{ Therefore, }\left|\begin{array}{ccc}5-\alpha& 0-0 & 0-0 \\ 0 & 3-\alpha& -2 \\ 0 & -1 & 2-\alpha \end{array}\right|$=0 \\ \Rightarrow \quad(5-\alpha)(3-\alpha)(2-\alpha)-2=0 \\ \Rightarrow \quad(5-\alpha)$\left(6-3 \alpha-2 \alpha+\alpha^{2}-2\right]$=0 \\ \Rightarrow \quad(5-\alpha) \left(\alpha^{2}-5 \alpha+4\right]$=0 \\ \Rightarrow \quad(5-\alpha)(\alpha-1)(\alpha-4)=0 \\ \Rightarrow \quad \alpha=1,4,5 \end{array}