Answer: \(\mathbf p=3\mathbf i+\mathbf j+2\mathbf k\) and \(\mathbf q=\mathbf i+2\mathbf j-3\mathbf k\).

One suitable equation of \(\Pi\) is

\(\mathbf r=\begin{pmatrix}3\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}-22\\-19\\5\end{pmatrix}+\mu\begin{pmatrix}2\\-1\\5\end{pmatrix}.\)

The line \(l_3\) may be written as

\(\mathbf r=\begin{pmatrix}-1\\-1\\0\end{pmatrix}+\lambda\begin{pmatrix}4\\3\\0\end{pmatrix}.\)

Let

\(P=(6+3s,\,-3-4s,\,-2s),\qquad Q=(2+t,\,-1-3t,\,-4-t).\)

Then

\(\overrightarrow{PQ}=Q-P=(t-3s-4,\,2+4s-3t,\,-4-t+2s).\)

Since \(PQ\) is perpendicular to both direction vectors \((3,-4,-2)\) and \((1,-3,-1)\),

\(\overrightarrow{PQ}\cdot(3,-4,-2)=0,\qquad \overrightarrow{PQ}\cdot(1,-3,-1)=0.\)

These give

\(17t-29s-12=0,\qquad 11t-17s-6=0.\)

Solving gives \(s=-1\) and \(t=-1\). Therefore

\(P=(3,1,2),\qquad Q=(1,2,-3).\)

So the position vector of \(P\) is \(3\mathbf i+\mathbf j+2\mathbf k\), and the position vector of \(Q\) is \(\mathbf i+2\mathbf j-3\mathbf k\).

The plane \(\Pi\) is perpendicular to \(l_1\), so its normal vector is the direction vector of \(l_1\), namely \((3,-4,-2)\). We need two independent vectors in the plane, each perpendicular to \((3,-4,-2)\). Two such vectors are

\(\begin{pmatrix}2\\-1\\5\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}-22\\-19\\5\end{pmatrix}.\)

Hence one equation of \(\Pi\) is

\(\mathbf r=\begin{pmatrix}3\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}-22\\-19\\5\end{pmatrix}+\mu\begin{pmatrix}2\\-1\\5\end{pmatrix}.\)

Equivalently, since \(\Pi\) has normal \((3,-4,-2)\) and passes through \((3,1,2)\), its cartesian equation is

\(3(x-3)-4(y-1)-2(z-2)=0,\)

so

\(3x-4y-2z=1.\)

The plane \(\mathbf r=p\mathbf i+q\mathbf j\) is the plane \(z=0\). Intersecting with \(\Pi\), we get

\(3x-4y=1.\)

One point on this line is \((-1,-1,0)\), and a direction vector is \((4,3,0)\). Therefore

\(\mathbf r=\begin{pmatrix}-1\\-1\\0\end{pmatrix}+\lambda\begin{pmatrix}4\\3\\0\end{pmatrix}.\)