Solution :

Vector equation of the plane passing through point $\overrightarrow{a}$ and normal to $\overrightarrow{n}$ is

$(r-\overrightarrow{a})=0$

$\overrightarrow{r} - \overrightarrow{n} = \overrightarrow{a} . \overrightarrow{n} $

Here $\overrightarrow{a}= 2 \hat i - \hat k$

Hence the vector equation of the plane is

$\overrightarrow{r}. (2 \hat i - 3\hat j +3 \hat k) =(2 \hat i - \hat k) .(2 \hat i - 3 \hat j +3 \hat k)$

((ie) $\overrightarrow{r} .(2 \hat i - 3 \hat j +3 \hat k) =4-3=1$

Hence $\overrightarrow{r}( 2\hat i - 3\hat j +3 \hat k) = 4-3 =1$

Hence $\overrightarrow{r} (2 \hat i -3 \hat j +3 \hat)=1$ is the required vector equation of the plane .

The Cartesian equation is

$(a \hat i + y \hat j +z \hat k) (2 \hat i - 3\hat j +3 \hat k) =1$

(ie) $2x-3y+3z =1$ is the required equation of the plane .