Let $ \mathbf{a} = \begin{pmatrix} -2 \\ 3 \\ 5 \end{pmatrix}, $ $ \mathbf{b} = \begin{pmatrix} 7 \\ 0 \\ -1 \end{pmatrix}, $ $ \mathbf{c} = \begin{pmatrix} -3 \\ -2 \\ -5 \end{pmatrix}, $ and $ \mathbf{d} = \begin{pmatrix} 3 \\ 4 \\ 7 \end{pmatrix} $. Then line $ AB $ is parameterized by \[\mathbf{a} + t (\mathbf{b} - \mathbf{a}) = \begin{pmatrix} -2 + 9t \\ 3 - 3t \\ 5 - 6t \end{pmatrix}.\]Also, line $ CD $ is parameterized by \[\mathbf{c} + s (\mathbf{d} - \mathbf{c}) = \begin{pmatrix} -3 + 6s \\ -2 + 6s \\ -5 + 12s \end{pmatrix}.\]Thus, we want \begin{align*} -2 + 9t &= -3 + 6s, \\ 3 - 3t &= -2 + 6s, \\ 5 - 6t &= -5 + 12s.\end{align*}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)} $.