(a) The normal vector to the plane \(\Pi\) is found using the cross product of the direction vectors \(\mathbf{i} + \mathbf{k}\) and \(2\mathbf{i} + 3\mathbf{j}\). The determinant is:

This gives the normal vector \(\begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}\).

Substitute the point \((-2, 3, 3)\) into the plane equation:

\(-3(-2) + 2(3) + 3(3) = 21\)

Thus, the Cartesian equation is \(-3x + 2y + 3z = 21\).

(b) The line \(l\) has the equation \(\begin{pmatrix} 2 \\ -3 \\ 5 + t \end{pmatrix}\). Substitute into the plane equation:

\(-3(2) + 2(-3) + 3(5 + t) = 21\)

This simplifies to \(t = 6\), giving the point \(\begin{pmatrix} 2 \\ -3 \\ 11 \end{pmatrix}\).

(c) The direction vector of \(l\) is \(\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\). The normal vector to \(\Pi\) is \(\begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}\). The dot product is:

\(0 \cdot (-3) + 0 \cdot 2 + 1 \cdot 3 = 3\)

The magnitudes are \(\sqrt{1}\) and \(\sqrt{22}\), so \(\cos \alpha = \frac{3}{\sqrt{22}}\).

The acute angle is \(90^\circ - \alpha = 39.8^\circ\).

(d) The direction vector from \(P\) to the plane is \(\begin{pmatrix} 0 \\ -3 \\ 7 \end{pmatrix}\). The dot product with the normal vector is:

\(\frac{1}{\sqrt{22}} \begin{pmatrix} -2 \\ 3 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix} = \frac{18}{\sqrt{22}} = 3.84\).