Normal of the plane $P_1$ determined by the vectors $\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$,
$$
\mathrm{n}_1=[(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \times(3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})]=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
0 & 1 & -1 \\
0 & 3 & -2
\end{array}\right|=\hat{\mathbf{i}}
$$
Normal of the plane $P_2$ determined by the vectors $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ and $\hat{\mathbf{i}}-3 \hat{\mathbf{j}}$,
$$
\begin{aligned}
& \mathrm{n}_2=[(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \times(\hat{\mathbf{i}}-3 \hat{\mathbf{j}})] \\
& =\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
2 & 3 & 0 \\
1 & -3 & 0
\end{array}\right|=\hat{\mathbf{i}}(0)-\hat{\mathbf{j}}(0)+\hat{\mathbf{k}}(-6-3)=-9 \hat{\mathbf{k}}
\end{aligned}
$$
Since, $a$ is parallel to the line of intersection of planes $P_1$ and $P_2$.
$$
a= \pm\left(n_1 \times n_2\right) \quad= \pm\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 0 & 0 \\
0 & 0 & -9
\end{array}\right|= \pm 9 \hat{\mathbf{j}}
$$
Now, angle $\theta$ between $9 \hat{\mathbf{j}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is
$$
\begin{aligned}
& \cos \theta= \pm\left(\frac{9}{9 \sqrt{3}}\right)= \pm\left(\frac{1}{\sqrt{3}}\right) \\
\therefore \quad & \theta=\cos \left( \pm \frac{}{\sqrt{ }}\right)
\end{aligned}
$$