(i) The direction vector of line AB is \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{i} + 4\mathbf{j}) - (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\).

Thus, the vector equation of the line AB is \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda(2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})\).

(ii) Since OP is perpendicular to AB, the dot product \(\overrightarrow{OP} \cdot \overrightarrow{AB} = 0\).

Let \(\overrightarrow{OP} = (1 + 2\lambda)\mathbf{i} + (2 + 2\lambda)\mathbf{j} + (2 - 2\lambda)\mathbf{k}\).

Then, \(\overrightarrow{OP} \cdot \overrightarrow{AB} = (1 + 2\lambda)(2) + (2 + 2\lambda)(2) + (2 - 2\lambda)(-2) = 0\).

Simplifying, \(2 + 4\lambda + 4 + 4\lambda - 4 + 4\lambda = 0\).

Thus, \(12\lambda + 2 = 0\), giving \(\lambda = -\frac{1}{6}\).

Substitute \(\lambda = -\frac{1}{6}\) into \(\overrightarrow{OP}\):

\(\overrightarrow{OP} = (1 + 2(-\frac{1}{6}))\mathbf{i} + (2 + 2(-\frac{1}{6}))\mathbf{j} + (2 - 2(-\frac{1}{6}))\mathbf{k}\).

\(\overrightarrow{OP} = \frac{2}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} + \frac{7}{3}\mathbf{k}\).