Given, the 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.

Apply the scalar triple product of $$(2 \hat{i}+k \hat{j}+2 \hat{k}),(5 \hat{i}+2 \hat{j}+k \hat{k})$$ and $$(-2 \hat{i}+0 \hat{j}+0 \hat{k})$$ is zero.

$$\begin{aligned} & \Rightarrow\left|\begin{array}{ccc} -2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{array}\right|=0 \\ & \Rightarrow-2\left(k^2-4\right)=0 \\ & \Rightarrow \quad k= \pm 2 \\ \end{aligned}$$

For the equation of plane containing given lines, apply scalar triple product of $$(x-1) \hat{i}+(y+1) \hat{j}+z \hat{k}, 2 \hat{i}$$ and $$2 \hat{i}+k \hat{j}+2 \hat{k}$$ is zero.

$$\begin{array}{lrl} \Rightarrow & \left|\begin{array}{ccc} x-1 & y+1 & z \\ -2 & 0 & 0 \\ 2 & \pm 2 & 2 \end{array}\right|=0 \\ \Rightarrow & 4(y+1)-2 k z=0 \\ \Rightarrow & 2(y+1)-( \pm 2) z=0 \\ \Rightarrow & y+1 \mp z=0 \\ \Rightarrow & y-z+1=0, y+z+1=0 \end{array}$$