The vector equation of the plane containing the line $\vec{r} = (-2\hat{i}-3\hat{j} + 4\hat{k}) +λ(3\hat{i}-2\hat{j} -\hat{k})$ and the point  $ \hat{i}+2\hat{j} + 3\hat{k}$, is

$\vec{r}. (\hat{i} + 3\hat{k}) = 10 $

The required plane passes through the points $P(-2\hat{i}-3\hat{j} + 4\hat{k})$and $Q(\hat{i}+2\hat{j} +3\hat{k})$ and iots parallel to the vector $\vec{b}= 3\hat{i} - 2\hat{j} - \hat{k}.$

So, it is normal to the vector

$\vec{n} = \vec{PQ}× \vec{b}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\-3 & -5 & 1\\3 & -2 & -1\end{vmatrix}= 7\hat{i} + 21 \hat{k}$

So, equation of the plane is 

$\begin{Bmatrix}\vec{r}-(-2\hat{i}-3\hat{j} + 4\hat{k})\end{Bmatrix}. (7\hat{i}+21\hat{k})= 0 ⇒ \vec{r}. (\hat{i}+3\hat{k})=10$