(a) The direction vectors of \(l_1\) and \(l_2\) are \(\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\) respectively. The vector between points on \(l_1\) and \(l_2\) is \(\begin{pmatrix} -4 \\ 0 \\ 1 \end{pmatrix}\). The cross product of the direction vectors is \(\begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). The shortest distance is given by \(\frac{1}{\sqrt{30}} \begin{vmatrix} -4 & 0 & 1 \\ -2 & -2 & -1 \end{vmatrix} = \frac{21}{\sqrt{30}}\).

(b) The normal to \(\Pi_1\) is the cross product of the direction vector of \(l_1\) and the direction vector of \(l_2\), which is \(\begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). Using the point \((1, 4, -1)\) on \(l_1\), the equation of \(\Pi_1\) is \(5x - 2y - z = -2\).

(c) The direction vector of the line of intersection is \(\begin{pmatrix} 1 \\ 4 \\ -3 \end{pmatrix}\). The normal to \(\Pi_2\) is \(\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}\). Using the point \((1, 1, 1)\), the equation of \(\Pi_2\) is \(x + 2y + 3z = 6\).