Answer: \(P=(1,-1,4)\).

\(\mathbf n=\begin{pmatrix}1\\2\\1\end{pmatrix}\), and \(\mathbf w=\begin{pmatrix}-5\\4\\-3\end{pmatrix}\).

An equation of \(PQ\) is \(\mathbf r=\begin{pmatrix}1\\-1\\4\end{pmatrix}+\mu\begin{pmatrix}5\\1\\-7\end{pmatrix}\).

The line has vector equation

\(\mathbf r=\begin{pmatrix}3\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-2\end{pmatrix}\),

so a general point on it is \((3+2\lambda,\lambda,2-2\lambda)\).

Substitute this into the plane equation \(\mathbf r\cdot(1,2,1)=3\):

\((3+2\lambda)+2\lambda+(2-2\lambda)=3\).

Thus \(5+2\lambda=3\), so \(\lambda=-1\). Therefore

\(P=(1,-1,4)\).

A vector perpendicular to \(\Pi\) is the normal vector

\(\mathbf n=\begin{pmatrix}1\\2\\1\end{pmatrix}\).

Now

\(\mathbf w=\mathbf n\times\begin{pmatrix}2\\1\\-2\end{pmatrix}=\begin{pmatrix}1\\2\\1\end{pmatrix}\times\begin{pmatrix}2\\1\\-2\end{pmatrix}=\begin{pmatrix}-5\\4\\-3\end{pmatrix}\).

The line \(PQ\) lies in the plane and is perpendicular to \(\mathbf w\), so its direction may be found from

\(\mathbf w\times\mathbf n=\begin{pmatrix}-5\\4\\-3\end{pmatrix}\times\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}10\\2\\-14\end{pmatrix}\).

This is parallel to \(\begin{pmatrix}5\\1\\-7\end{pmatrix}\). Since \(PQ\) passes through \(P\),

\(\mathbf r=\begin{pmatrix}1\\-1\\4\end{pmatrix}+\mu\begin{pmatrix}5\\1\\-7\end{pmatrix}\).