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) $\text{Given: } L_1:\ x = 5,\ \dfrac{y}{3-\alpha} = \dfrac{z}{-2}$ $L_2:\ x = \alpha,\ \dfrac{y}{-1} = \dfrac{z}{2-\alpha}$ $\text{To find: } \alpha \text{ such that } L_1, L_2 \text{ are coplanar}$ $1.\ \text{point and direction of } L_1, L_2$ $2.\ \text{coplanar} \Leftrightarrow [\vec{A_2}-\vec{A_1},\ \vec{d_1},\ \vec{d_2}] = 0$ $3.\ \text{solve for } \alpha, \text{ then check the options}$ $L_1:\ A_1 = (5, 0, 0), \quad \vec{d_1} = (0,\ 3-\alpha,\ -2)$ $L_2:\ A_2 = (\alpha, 0, 0), \quad \vec{d_2} = (0,\ -1,\ 2-\alpha)$ $\text{Coplanar} \Leftrightarrow [\vec{A_2} - \vec{A_1},\ \vec{d_1},\ \vec{d_2}] = 0$ $\Leftrightarrow (\vec{A_2} - \vec{A_1}) \cdot (\vec{d_1} \times \vec{d_2}) = 0$ $\vec{A_2} - \vec{A_1} = (\alpha - 5,\ 0,\ 0)$ $\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[2pt] 0 & 3-\alpha & -2 \\[2pt] 0 & -1 & 2-\alpha \end{vmatrix}$ $= \left[(3-\alpha)(2-\alpha) - 2\right]\hat{i} = (\alpha^{2} - 5\alpha + 4)\,\hat{i}$ $\alpha^{2} - 5\alpha + 4 = (\alpha - 1)(\alpha - 4)$ $\vec{d_1} \times \vec{d_2} = (\alpha-1)(\alpha-4)\,\hat{i}$ $(\vec{A_2} - \vec{A_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (\alpha-5)(\alpha-1)(\alpha-4)$ $(\alpha-5)(\alpha-1)(\alpha-4) = 0 \Rightarrow \alpha = 1,\ 4,\ 5$ $\text{Given options are } 1, 2, 3, 4 \Rightarrow \alpha = 1 \text{ or } \alpha = 4$ $\alpha = 1:\ \vec{d_1} = (0, 2, -2),\ \vec{d_2} = (0, -1, 1) \Rightarrow \vec{d_1} \parallel \vec{d_2}$ $\alpha = 4:\ \vec{d_1} = \vec{d_2} = (0, -1, -2) \Rightarrow \vec{d_1} \parallel \vec{d_2}$ $\alpha = 2:\ (\alpha-5)(\alpha-1)(\alpha-4) = (-3)(1)(-2) = 6 \neq 0$ $\alpha = 3:\ (-2)(2)(-1) = 4 \neq 0$ $\boxed{\text{Answer: } \alpha = 1 \text{ and } \alpha = 4 \Rightarrow \text{options (1) and (4)}}$