(a) The normal vector to \(\Pi_1\) can be found using the cross product of the direction vectors \(\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}\) and \(3\mathbf{i} - \mathbf{k}\). This gives \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -3 \\ 3 & 0 & -1 \end{vmatrix} = \mathbf{i}(2 - 0) - \mathbf{j}(-1 - (-9)) + \mathbf{k}(0 - (-6)) = 2\mathbf{i} - 8\mathbf{j} + 6\mathbf{k}\).

Using the point \((1, -1, -2)\), the equation is \(1(x - 1) - 4(y + 1) + 3(z + 2) = 0\), simplifying to \(x - 4y + 3z = -1\).

(b) The direction vector of \(l\) is \(\mathbf{i} + \mathbf{j} + \mathbf{k}\). The dot product with the normal vector \(1\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}\) is \(1(1) - 4(1) + 3(1) = 0\), showing \(l\) is parallel to \(\Pi_1\).

(c) The distance from a point on \(l\) to \(\Pi_1\) is \(\frac{|(-3)(1) + 0(-4) + 1(3) + 1|}{\sqrt{1^2 + (-4)^2 + 3^2}} = \frac{1}{\sqrt{26}}\).

(d) A point common to both planes is found by solving \(x - 4y + 3z = -1\) and \(3x + 3y + 2z = 1\). Solving these gives a point \(\begin{pmatrix} \frac{5}{7} \\ 0 \\ -\frac{4}{7} \end{pmatrix}\). The direction vector of the line of intersection is the cross product of the normals \(\begin{pmatrix} 1 \\ -4 \\ 3 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}\), which is \(\begin{pmatrix} -17 \\ 7 \\ 15 \end{pmatrix}\). Thus, the line is \(\mathbf{r} = \begin{pmatrix} \frac{5}{7} \\ 0 \\ -\frac{4}{7} \end{pmatrix} + \lambda \begin{pmatrix} -17 \\ 7 \\ 15 \end{pmatrix}\).