(a) The direction vector of line AB is \(\mathbf{AB} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}\).

The direction vector of line CD is \(\mathbf{CD} = \mathbf{j} + (\lambda - 3)\mathbf{k}\).

Find the common perpendicular using the cross product and the formula for perpendicular distance:

\(\frac{1}{\sqrt{(\lambda - 2)^2 + (4\lambda - 12)^2 + 16}} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ 0 & 1 & \lambda - 3 \end{vmatrix} = 3\)

Solving gives \(\lambda^2 - 5\lambda + 4 = 0\).

(b)(i) The equation of \(\Pi_1\) is \(\mathbf{r} = 7\mathbf{i} + 4\mathbf{j} - \mathbf{k} + s(4\mathbf{i} - \mathbf{j} + \mathbf{k}) + t(-5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})\).

(b)(ii) The normal to \(\Pi_2\) is found using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ -5 & 3 & 7 \end{vmatrix} = -8\mathbf{i} - 25\mathbf{j} + 7\mathbf{k}\)

Substitute a point to find \(-8x - 25y + 7z = -163\).

(c) The angle between \(\Pi_1\) and \(\Pi_2\) is found using the dot product of normals:

\(\cos \theta = \frac{414}{\sqrt{243 \times 738}}\)

\(\theta = 12.1^\circ\).