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)

Two lines L₁ : x=5, y/(3-α)=z/(-2) and L₂ : x=α, y/(-1)=z/(2-α) are coplanar. Then α can take value(s) $L_1 : x=5,\ \dfrac{y}{3-\alpha}=\dfrac{z}{-2}$ $L_2 : x=\alpha,\ \dfrac{y}{-1}=\dfrac{z}{2-\alpha}$ $\text{Given: } L_1, L_2 \text{ are coplanar}; \quad \text{To find: values of } \alpha$ $L_1: \dfrac{x-5}{0} = \dfrac{y}{3-\alpha} = \dfrac{z}{-2}$ $A(5,0,0), \quad \vec{d_1} = (0,\ 3-\alpha,\ -2)$ $L_2: \dfrac{x-\alpha}{0} = \dfrac{y}{-1} = \dfrac{z}{2-\alpha}$ $B(\alpha,0,0), \quad \vec{d_2} = (0,\ -1,\ 2-\alpha)$ $\text{Coplanar} \iff \overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 0$ $\text{Plan: } 1.\ \overrightarrow{AB} = B - A$ $2.\ \vec{d_1} \times \vec{d_2}$ $3.\ \overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 0$ $\overrightarrow{AB} = B - A = (\alpha-5,\ 0,\ 0)$ $\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3-\alpha & -2 \\ 0 & -1 & 2-\alpha \end{vmatrix}$ $= \big[(3-\alpha)(2-\alpha) - 2\big]\,\hat{i}$ $\overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2}) = (\alpha-5)\big[(3-\alpha)(2-\alpha) - 2\big] = 0$ $\alpha - 5 = 0 \ \Rightarrow\ \alpha = 5$ $\text{(the two lines meet at } (5,0,0)\text{)}$ $(3-\alpha)(2-\alpha) = 2 \Rightarrow \alpha^{2} - 5\alpha + 4 = 0$ $(\alpha-1)(\alpha-4) = 0 \Rightarrow \alpha = 1 \ \text{or}\ \alpha = 4$ $\alpha \in \{1,\ 4,\ 5\}$ $\text{Correct options: } (1)\ \text{and}\ (4)$