Find the intersection point of the lines defined by
\[\begin{pmatrix} 4 \\ -7 \\ 0 \end{pmatrix} + t \begin{pmatrix} -3 \\ 3 \\ 1 \end{pmatrix}\]and
\[\begin{pmatrix} -2 \\ -5 \\ 4/3 \end{pmatrix} + u \begin{pmatrix} 2 \\ 4 \\ 1/3 \end{pmatrix}.\]
Setting the coordinates to be equal, we obtain the system of equations
\begin{align*}
4 - 3t &= -2 + 2u, \\
-7 + 3t &= -5 + 4u, \\
t &= \frac{4}{3} + \frac{1}{3} u.
\end{align*}Solving this system, we find $t = \frac{14}{9}$ and $u = \frac{2}{3}.$  Hence, the point of intersection is $\boxed{\left( -\frac{2}{3}, -\frac{7}{3}, \frac{14}{9} \right)}.$