The vector equation of the plane passing through the points $\hat{i} + \hat{j} - 2 \hat{k}, 2\hat{i} - \hat{j} + \hat{k}$ and $ \hat{i} + 2\hat{j} +\hat{k},$ is

$\vec{r}. (9\hat{i}+3\hat{j}-\hat{k})=14$

Let A, B, C be the points with position vectors $\hat{i}+\hat{j}-2\hat{k}, 2\hat{i}-\hat{j}+\hat{k}$ and $ \hat{i}+2\hat{j}+\hat{k}$ respectively. Then, 

$\vec{AB}= \hat{i}-2\hat{j}+3\hat{k}$ and, $\vec{AB}=-\hat{i}+3\hat{j}+0\hat{k}$

A vector normal to the plane containing points A, B, and C, is

$\vec{n} = \vec{AB} × \vec{AC}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & -2 & 3\\-1 & 3 & 0\end{vmatrix}= -9\hat{i}-3\hat{j}+k\hat{k}$

The required plane passes through the point having position vector $\vec{a} = \hat{i}+\hat{j}-2\hat{k}$ and is normal to the vector $ \vec{n} = - 9 \hat{i} -3\hat{j}+\hat{k}.$

So, its vector equation is

$⇒\vec{r}.\vec{n}= \vec{a}.\vec{n}⇒\vec{i}.(9\hat{i}+3\hat{j}-\hat{k})=14$