Vector equation of the line passing through $(-2,-1,-3) \&(1,-3,3)$ is:
$$
\overrightarrow{\mathrm{N}}=(-2-1) \hat{\mathrm{i}}+(-1+3) \hat{\mathrm{j}}+(-3-3) \hat{\mathrm{k}} \text {. }
$$
or $\overrightarrow{\mathrm{N}}=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}$.
And this line meets the plane perpendicularly at $(1,-3,3)$ with position vector $=\vec{a}=\hat{i}-3 \hat{j}+3 \hat{k}$.
Therefore the equation of plane is:
$$
\begin{aligned}
&(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}}) \cdot \mathrm{N}=0 \\
&\Rightarrow[(x \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})-(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})] \cdot(-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})=0
\end{aligned}
$$
$\Rightarrow[(x-1) \hat{\mathrm{i}}+(\mathrm{y}+3) \hat{\mathrm{j}}+(\mathrm{z}-3) \hat{\mathrm{k}}] \cdot(-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})=0$
$\Rightarrow-3 \mathrm{x}+3+2 \mathrm{y}+6-6 \mathrm{z}+18=0$
$\therefore 3 \mathrm{x}-2 \mathrm{y}+6 \mathrm{z}-27=0$