(a) To find the shortest distance between \(l_1\) and \(l_2\), we first find the direction vector from one line to another: \(\begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} - \begin{pmatrix} -2 \\ -3 \\ -5 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix}\).

Next, find the common perpendicular by taking the cross product of the direction vectors of \(l_1\) and \(l_2\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 5 \\ 2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} 18 \\ 14 \\ 6 \end{pmatrix} \sim \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix}\).

Use the formula for shortest distance:

\(\frac{1}{\sqrt{139}} \left| \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix} \right| = \frac{67}{\sqrt{139}} \approx 5.68\).

(b) To find the equation of the plane \(\Pi\), find a vector perpendicular to the plane using the cross product of the direction vector of \(l_1\) and the vector from the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\) to a point on \(l_1\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -4 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 3 \\ -3 \\ -1 \end{pmatrix} \sim \begin{pmatrix} 9 \\ -3 \\ -1 \end{pmatrix}\).

Use the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\) in the plane equation:

\(1(-1) + 3(-3) - 1(-4) = -6 \Rightarrow x + 3y - z = -6\).