Given vectors
$$
\begin{aligned}
& \mathbf{O A} & =2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\text { and } & \mathbf{O B} & =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\
\because & |\mathbf{O A}| & =\sqrt{4+4+1}=3=m \text { (let) and } \\
& |\mathbf{O B}| & =\sqrt{4+16+16}=6=n \text { (let) }
\end{aligned}
$$
and
$\because$ The angle bisector of $\angle B O A$ intersect the side $A B$ at point $P$ in the ratio $3: 6=1: 2$, so
$$
\begin{aligned}
\mathbf{O P} & =\frac{2(\mathbf{O A})+1(\mathbf{O B})}{3} \\
& =\frac{6 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}}{3}=2 \hat{\mathbf{i}}+\frac{8}{3} \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
\therefore \quad|\mathbf{O P}| & =\sqrt{2^2+\left(\frac{8}{3}\right)^2+2^2} \\
& =\sqrt{4+\frac{64}{9}+4} \\
& =\sqrt{\frac{72+64}{9}}=\sqrt{\frac{136}{9}}=k \\
\therefore \quad 9 k^2 & =9\left(\frac{136}{9}\right)=136
\end{aligned}
$$