The plane containing the line of intersection of the planes $\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})=0 \quad$ and $\overrightarrow{\mathrm{r}} .(\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=0$ is
$\begin{array}{l}
\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})+\lambda[\hat{\mathrm{r}} \cdot(\hat{\mathrm{j}}+2 \hat{\mathrm{k}})]=0 \\
\Rightarrow \overrightarrow{\mathrm{r}} \cdot[\hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2 \lambda-1) \hat{\mathrm{k}}]=0 ...(1)
\end{array}$
Since the plane (i) passes through the point $(-1,-1,-1)$ or $(-\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})$, so this point must satisfy (i). Hence,
$\begin{array}{l}
(-\hat{i}-\hat{j}-\hat{k}) \cdot[\hat{i}+(3 \times \lambda) \hat{j}+(2 \lambda-1) \hat{k}]=0 \\
\Rightarrow-1-(3+\lambda)-(2 \lambda-1)=0 \\
\Rightarrow-3 \lambda-3=0 \Rightarrow \lambda=-1
\end{array}$
Substituting this value for $\lambda$ in (i), we get the required plane $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=0$