If the straight lines $\frac{x-1}{2} = \frac{y+1}{k} = \frac{z}{2}$ and $\frac{x+1}{5} = \frac{y+1}{2} = \frac{z}{k}$ are coplanar, then the plane (s) containing these two lines is (are)

$L_1: \dfrac{x-1}{2} = \dfrac{y+1}{k} = \dfrac{z}{2}$ $L_2: \dfrac{x+1}{5} = \dfrac{y+1}{2} = \dfrac{z}{k}$ $\text{Given: } L_1, L_2 \text{ are coplanar}$ $\text{To find: plane(s) containing both lines}$ $L_1: A(1, -1, 0), \quad \vec{a} = (2, k, 2)$ $L_2: B(-1, -1, 0), \quad \vec{b} = (5, 2, k)$ $\text{Coplanar} \Leftrightarrow \vec{AB} \cdot (\vec{a} \times \vec{b}) = 0$ $\vec{AB} = B - A = (-2, 0, 0)$ $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & k & 2 \\ 5 & 2 & k \end{vmatrix}$ $= (k^2 - 4,\ 10 - 2k,\ 4 - 5k)$ $\vec{AB}\cdot(\vec{a}\times\vec{b}) = (-2)(k^2 - 4) = 0$ $k^2 = 4 \Rightarrow k = \pm 2$ $k = 2:\ \vec{a} = (2,2,2),\ \vec{b} = (5,2,2)$ $\vec{a}\times\vec{b} = (0, 6, -6) \Rightarrow \vec{n} = (0, 1, -1)$ $\text{Plane: } y - z = -1 \quad \Rightarrow \text{option (3)}$ $k = -2:\ \vec{a} = (2,-2,2),\ \vec{b} = (5,2,-2)$ $\vec{a}\times\vec{b} = (0, 14, 14) \Rightarrow \vec{n} = (0, 1, 1)$ $\text{Plane: } y + z = -1 \quad \Rightarrow \text{option (2)}$ $\boxed{y + z = -1\ (2), \quad y - z = -1\ (3)}$