A plane $P$ is parmeterized by
\[\mathbf{v} = \begin{pmatrix} 1 \\ 6 \\ 7 \end{pmatrix} + t \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix} + s \begin{pmatrix} 2 \\ -3 \\ -5 \end{pmatrix},\]and line $L$ is parameterized by
\[\mathbf{w} = \begin{pmatrix} 7 \\ 4 \\ 1 \end{pmatrix} + u \begin{pmatrix} 3 \\ 0 \\ -1 \end{pmatrix}.\]Find the intersection of plane $P$ and line $L.$
Setting the parameterizations to be equal, we obtain
\begin{align*}
1 + 2t + 2s &= 7 + 3u, \\
6 - t - 3s &= 4 \\
7 - t - 5s &= 1 - u.
\end{align*}Solving this system, we find $s = 1,$ $t = -1,$ and $u = -2.$  Thus, the point of intersection is $\boxed{\begin{pmatrix} 1 \\ 4 \\ 3 \end{pmatrix}}.$