The vectors 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 two given lines pass through the point having position vector $\vec{a}=\hat{i}+\hat{j}$ and are parallel to the vectors $\vec{b_1}= \hat{i} + 2\hat{j}-\hat{k}$ and $\vec{b_2}= -\hat{i} + 2\hat{j}-2\hat{k}$ respectively. Therefore, the plane containing the given lines also passes through the point with position vector $\vec{a}= \hat{i} + \hat{j}$. Since the plane contains the lines which are parallel to the vectors $\vec{b_1}$ and $\vec{b_2}$ respectively. Therefore, the plane is normal to the vector $\vec{n}$ given by

$\vec{n} = \vec{b_1}× \vec{b_2}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & 2 & -1\\-1 & 1 & -2\end{vmatrix}= -3\hat{i} + 3\hat{j}+3\hat{k}$

Thus, the vector equation of the required planes 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} .(-\hat{i} + \hat{j}+\hat{k})=0 $