(a) To find the Cartesian equation of the plane \(\Pi\), we need a normal vector. The direction vectors are \(\mathbf{i} - 2\mathbf{j} - \mathbf{k}\) and \(3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\). The normal vector is the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -1 \\ 3 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

Using the point \((2, 3, -2)\) on the plane, substitute into \(6x - y + 8z = d\):

\(6(2) - 3 + 8(-2) = -7\)

Thus, the equation is \(6x - y + 8z = -7\).

(b) The position vector of the foot of the perpendicular \(\mathbf{F}\) is given by:

\(\mathbf{OF} = \mathbf{OP} + t \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

\(\mathbf{OF} = \begin{pmatrix} 4 \\ 2 \\ 9 \end{pmatrix} + t \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix} = \begin{pmatrix} 4 + 6t \\ 2 - t \\ 9 + 8t \end{pmatrix}\)

Substitute into the plane equation:

\(6(4 + 6t) - (2 - t) + 8(9 + 8t) = -7\)

\(101t + 94 = -7\)

\(t = -1\)

\(\mathbf{OF} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\)

(c) The acute angle \(\alpha\) between \(l\) and \(\Pi\) is found using:

\(\cos \alpha = \frac{\mathbf{d} \cdot \mathbf{n}}{\|\mathbf{d}\| \|\mathbf{n}\|}\)

\(\mathbf{d} = \begin{pmatrix} 3 \\ 5 \\ -1 \end{pmatrix}, \mathbf{n} = \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

\(\cos \alpha = \frac{5}{\sqrt{35} \sqrt{101}}\)

\(\alpha = 4.8^\circ\)