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)

$L_1: \dfrac{x-1}{2}=\dfrac{y+1}{k}=\dfrac{z}{2}$ $L_2: \dfrac{x+1}{5}=\dfrac{y+1}{2}=\dfrac{z}{k}$ $L_1, L_2 \text{ are coplanar}$ $\text{To find: plane(s) containing both } L_1 \text{ and } L_2$ $\text{Two lines coplanar} \iff \vec{AB}\cdot(\vec{d_1}\times\vec{d_2})=0$ $\text{Plan: } 1.\ A,\ \vec{d_1},\ B,\ \vec{d_2}$ $2.\ \vec{AB}\cdot(\vec{d_1}\times\vec{d_2})=0 \Rightarrow k$ $3.\ \text{normal} = \vec{d_1}\times\vec{d_2} \Rightarrow \text{plane through } A$ $A(1,-1,0),\quad \vec{d_1}=(2,k,2)$ $B(-1,-1,0),\quad \vec{d_2}=(5,2,k)$ $\vec{AB}=B-A=(-1-1,\ -1-(-1),\ 0-0)$ $\vec{AB}=(-2,0,0)$ $\vec{d_1}\times\vec{d_2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&k&2\\5&2&k\end{vmatrix}$ $=\hat{i}(k^2-4)-\hat{j}(2k-10)+\hat{k}(4-5k)$ $\vec{d_1}\times\vec{d_2}=(k^2-4,\ 10-2k,\ 4-5k)$ $\vec{AB}\cdot(\vec{d_1}\times\vec{d_2})=-2(k^2-4)=0$ $k^2=4\ \Rightarrow\ k=\pm 2$ $k=2:\ (k^2-4,\ 10-2k,\ 4-5k)=(0,\ 6,\ -6)$ $\text{normal} \parallel (0,1,-1)$ $y-z=c,\quad A(1,-1,0)\Rightarrow c=-1-0=-1$ $y-z=-1\quad \text{option (3)}$ $k=-2:\ (k^2-4,\ 10-2k,\ 4-5k)=(0,\ 14,\ 14)$ $\text{normal} \parallel (0,1,1)$ $y+z=c,\quad A(1,-1,0)\Rightarrow c=-1+0=-1$ $y+z=-1\quad \text{option (2)}$ $B(-1,-1,0):\ -1-0=-1\ \checkmark,\quad -1+0=-1\ \checkmark$ $y+z=-1\ \text{ and }\ y-z=-1$ $\boxed{\text{options (2) and (3)}}$