Equation of the plane that contains the lines

$\vec{r}= (\hat{i} +\hat{j}) + λ(\hat{i} +2\hat{j}-\hat{k})$ and, $\vec{r} = (\hat{i} +\hat{j})+\mu (-\hat{i} +\hat{j}-2\hat{k})$ is

$\vec{r}.(-\hat{i} +\hat{j}+\hat{k})=0$

The lines are parallel to the vectors $\vec{b_1} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b_2}= -\hat{i} +\hat{j} - 2\hat{k}.$ Therefore, the plane is normal to the vector

$\vec{n} = \vec{b_1}×\vec{B_2}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & 2 & -1\\-1 & 1 & -2\end{vmatrix}= -3\hat{i} + 3\hat{j} + 3\hat{k}$

The required plane passes through $(\hat{i}-\hat{j})$ and is normal to the vector $\vec{n}$. Therefore, its equation is

$\vec{r}.\vec{n}=\vec{a}.\vec{n}$

$⇒\vec{r}. (-3\hat{i} + 3\hat{j} + 3\hat{k}) = (\hat{i} + \hat{j}_.(-3\hat{i} + 3\hat{j} + 3\hat{k})$

$⇒\vec{r}. (-3\hat{i} + 3\hat{j} + 3\hat{k})= -3 + 3⇒\vec{r}.(-\hat{i} +\hat{j}+\hat{k})=0$