(a) To find the equation of \(\Pi_1\), we need a normal vector to the plane. Since \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), the direction vector of \(l_2\), \(\begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}\), is parallel to \(\Pi_1\). The direction vector of \(l_1\) is \(\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\). The normal to \(\Pi_1\) is the cross product of these two vectors:
\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 1 & -4 & 2 \end{vmatrix} = \begin{pmatrix} 6 \\ -3 \\ 3 \end{pmatrix}\)
Substitute the point on \(l_1\), \((1, 3, -2)\), into the plane equation \(6x - 3y + 3z = d\) to find \(d\):
\(6(1) - 3(3) + 3(-2) = -5\)
Thus, the equation of \(\Pi_1\) is \(2x - y - 3z = 5\).
(b) The normal to \(\Pi_2\) is the direction vector of \(l_2\), \(\begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}\). The angle \(\theta\) between \(\Pi_1\) and \(\Pi_2\) is given by:
\(\cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\| \|\mathbf{n}_2\|} = \frac{\begin{pmatrix} 6 \\ -3 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}}{\sqrt{14} \sqrt{77}} = \frac{-7}{\sqrt{14 \times 77}}\)
\(\theta = 77.7^\circ\).
(c) To find \(\mathbf{PQ}\), we need points \(P\) and \(Q\) such that \(\mathbf{PQ}\) is perpendicular to both lines. Let \(\mathbf{OP} = \begin{pmatrix} 1 + 2\lambda \\ 3 + \lambda \\ -2 + \lambda \end{pmatrix}\) and \(\mathbf{OQ} = \begin{pmatrix} 1 + \mu \\ -2 - 4\mu \\ 9 + 2\mu \end{pmatrix}\).
\(\mathbf{PQ} = \mathbf{OQ} - \mathbf{OP} = \begin{pmatrix} \mu - 2\lambda \\ -5 - 4\mu - \lambda \\ 11 + 2\mu - \lambda \end{pmatrix}\).
Set \(\mathbf{PQ} \cdot \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = 0\) and \(\mathbf{PQ} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix} = 0\) to find \(\lambda\) and \(\mu\):
\(-6\lambda + 6 = 0 \rightarrow \lambda = 1\)
\(21\mu + 42 = 0 \rightarrow \mu = -2\)
Substitute back to find \(\mathbf{PQ}\):
\(\mathbf{r} = \begin{pmatrix} -1 \\ 6 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 6 \\ -3 \\ -9 \end{pmatrix}\).
