Given that,

Position vector of $A$ is $\hat{i}+2\hat{j}-3\hat{k}$ and it is parallel to vector $2\hat{i}+\hat{j}+2\hat{k}$.

The vector equation of line passing through the point $A$ and parallel to vector $\overrightarrow{b}$ is

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$

$\Rightarrow \overrightarrow{r}=\left(\hat{i}+2\hat{j}-3\hat{k}\right)+\lambda \left(2\hat{i}+\hat{j}+2\hat{k}\right)$

$\Rightarrow \overrightarrow{r}=\left(1+2\lambda \right)\hat{i}+\left(2+\lambda \right)\hat{j}+\left(-3+2\lambda \right)\hat{k} ...\left(1\right)$

The equation of plane $\mathrm{\pi }$ passing through the point $\hat{i}+\hat{j}+\hat{k}, \hat{i}-\hat{j}-\hat{k}$ and parallel to vector $\hat{i}-2\hat{j}$ is

$\overrightarrow{r}=\left(1-t\right)\overrightarrow{a}+t\overrightarrow{b}+s\overrightarrow{c}$

$=\left(1-t\right)\left(\hat{i}+\hat{j}+\hat{k}\right)+t\left(\hat{i}-\hat{j}-\hat{k}\right)+s(\left(\hat{i}-2\hat{j}\right)$

$=\left(1-t\right)\hat{i}+\left(1-t\right)\hat{j}+\left(1-t\right)\hat{k}+t\hat{i}-t\hat{j}-t\hat{k}+s\hat{i}-2s\hat{j}$

$=\left(1-t+t+s\right)\hat{i}+\left(1-t-t-2s\right)\hat{j}+\left(1-t-t+0\right)\hat{k}$

$=\left(1+s\right)\hat{i}+\left(1-2t-2s\right)\hat{j}+\left(1-2t\right)\hat{k} ...\left(2\right)$ 

From equations $\left(1\right) \& \left(2\right)$,

$1+2\lambda =1+s\Rightarrow 2\lambda =s$

$\Rightarrow \lambda =\frac{s}{2} ...\left(3\right)$

$\Rightarrow 2+\lambda =1-2t-2s$

$\Rightarrow \lambda +2t+2s=-1 ...\left(4\right)$

$-3+2\lambda =1-2t$

$\Rightarrow 2\lambda +2t=1+3$

$\Rightarrow 2\lambda +2t=4 ...\left(5\right)$

Solving equations $\left(4\right) \& \left(5\right)$,

$\left(4\right)-\left(5\right)\Rightarrow \lambda +2t+2s-2\lambda -2t=-1-4$

$\Rightarrow -\lambda +2s=-5$

$\Rightarrow -\frac{s}{2}+2s=-5$  [$\because$ From equation $\left(3\right)$]

$\Rightarrow -s+4s=-10$

$\Rightarrow 3s=-10$

$\Rightarrow s=\frac{-10}{3}$

$\therefore \lambda =\frac{s}{2}=\frac{-10}{6}$

From equation $\left(5\right)$,

$2\left(\frac{-10}{6}\right)+2t=4$

$\Rightarrow 2t=4+\frac{10}{3}$

$\Rightarrow 2t=\frac{22}{3}$

$\Rightarrow t=\frac{11}{3}$

$\therefore$ The point of intersection is,

$r=\hat{i}+2\hat{j}-3\hat{k}-\frac{10}{6}\left(2\hat{i}+\hat{j}+2\hat{k}\right)$

$=\hat{i}+2\hat{j}-3\hat{k}-\frac{10}{3}\hat{i}-\frac{5}{3}\hat{j}-\frac{10}{3}\hat{k}$

$=\left(1-\frac{10}{3}\right)\hat{i}+\left(2-\frac{5}{3}\right)\hat{j}+\left(-3-\frac{10}{3}\right)\hat{k}$

$=\frac{-7}{3}\hat{i}+\frac{1}{3}\hat{j}-\frac{19}{3}\hat{k}$

$=\frac{1}{3}\left[-7\hat{i}+\hat{j}-19\hat{k}\right]$

$\therefore$ The point where the plane $\mathrm{\pi }$ meets the line $L$ is

$\frac{1}{3}\left[-7\hat{i}+\hat{j}-19\hat{k}\right]$.