(a) The position vector of \(M\) is \(\mathbf{OM} = 4\mathbf{i} + 2\mathbf{j}\).

For \(N\), since \(DN = 3NC\), \(N\) divides \(CD\) in the ratio 3:1. The position vector of \(C\) is \(4\mathbf{j}\) and \(D\) is \(4\mathbf{k}\). Using the section formula, \(\mathbf{ON} = \frac{1}{4}(3 \times 4\mathbf{k} + 1 \times 4\mathbf{j}) = 3\mathbf{j} + \mathbf{k}\).

The direction vector \(\mathbf{MN} = \mathbf{ON} - \mathbf{OM} = (3\mathbf{j} + \mathbf{k}) - (4\mathbf{i} + 2\mathbf{j}) = -4\mathbf{i} + \mathbf{j} + \mathbf{k}\).

The vector equation of the line through \(M\) and \(N\) is \(\mathbf{r} = 3\mathbf{j} + \mathbf{k} + \lambda(4\mathbf{i} - \mathbf{j} - \mathbf{k})\).

(b) The direction vector of \(MN\) is \(-4\mathbf{i} + \mathbf{j} + \mathbf{k}\), with magnitude \(\sqrt{(-4)^2 + 1^2 + 1^2} = \sqrt{18}\).

The projection of \(\mathbf{OM}\) on \(\mathbf{MN}\) is \(\frac{\mathbf{OM} \cdot \mathbf{MN}}{\|\mathbf{MN}\|} = \frac{(4\mathbf{i} + 2\mathbf{j}) \cdot (-4\mathbf{i} + \mathbf{j} + \mathbf{k})}{\sqrt{18}} = \frac{-16 + 2}{\sqrt{18}} = \frac{-14}{\sqrt{18}}\).

The perpendicular distance is \(\sqrt{\|\mathbf{OM}\|^2 - \left(\frac{-14}{\sqrt{18}}\right)^2} = \sqrt{20 - \frac{196}{18}} = \frac{1}{3}\sqrt{82}\).