jmc

Let $\mathbf{a}=$ $\mathbf{b}=$ $\mathbf{c}=$ and $\mathbf{d}=$ Then line $AB$ is parameterized by $$\mathbf{a}+t(\mathbf{b}-\mathbf{a})=\left(\begin{array}{c}-2+9t\\ 3-3t\\ 5-6t\end{array}\right).$$Also, line $CD$ is parameterized by $$\mathbf{c}+s(\mathbf{d}-\mathbf{c})=\left(\begin{array}{c}-3+6s\\ -2+6s\\ -5+12s\end{array}\right).$$Thus, we want $$\begin{array}{rl}-2+9t& =-3+6s,\\ 3-3t& =-2+6s,\\ 5-6t& =-5+12s.\end{array}$$Solving this system, we find $t=\frac{1}{3}$ and $s=\frac{2}{3}.$ We can find the point of intersection as $\boxed{(1,2,3)}.$