The line l has equation \(\mathbf{r} = \mathbf{i} - 2\mathbf{j} - 3\mathbf{k} + \lambda\bigl(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\bigr)\). The points A and B have position vectors \(-2\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and \(3\mathbf{i} - \mathbf{j} + \mathbf{k}\) respectively.

(a) The direction vector of line l is \(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\). The magnitude of this vector is \(\sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{6}\). Therefore, a unit vector in the direction of l is \(\dfrac{1}{\sqrt{6}}\bigl(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\bigr)\).

(b) The vector from A to B is \((3\mathbf{i} - \mathbf{j} + \mathbf{k}) - (-2\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\). Thus, a vector equation for line m is \(\mathbf{r} = -2\mathbf{i} + 2\mathbf{j} - \mathbf{k} + \mu\bigl(5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\bigr)\).

(c) The direction vector of l is \(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and for m is \(5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\). These vectors are not scalar multiples of each other, so the lines are not parallel. Solving the system obtained by equating components of the general points on l and m does not yield a consistent solution, indicating the lines do not intersect. Therefore, lines l and m are skew.