(i) The vector \(\overrightarrow{AB}\) is given by:

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{i} + 4\mathbf{j}) - (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\)

Since \(\overrightarrow{AP} = \lambda \overrightarrow{AB}\), we have:

\(\overrightarrow{AP} = \lambda (2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = 2\lambda \mathbf{i} + 2\lambda \mathbf{j} - 2\lambda \mathbf{k}\)

Thus, \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) + (2\lambda \mathbf{i} + 2\lambda \mathbf{j} - 2\lambda \mathbf{k})\).

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

(ii) To find \(\lambda\) such that \(OP\) bisects \(\angle AOB\), equate the expressions for \(\cos AOP\) and \(\cos BOP\):

\(\frac{9 + 2\lambda}{3\sqrt{9 + 4\lambda + 12\lambda^2}} = \frac{11 + 14\lambda}{5\sqrt{9 + 4\lambda + 12\lambda^2}}\)

Solving this equation gives \(\lambda = \frac{3}{8}\).

(iii) Verify \(AP : PB = OA : OB\):

\(AP = \lambda \times AB = \frac{3}{8} \times \sqrt{12}\)

\(PB = (1 - \lambda) \times AB = \frac{5}{8} \times \sqrt{12}\)

\(OA = \sqrt{9}\), \(OB = \sqrt{25}\)

\(\frac{AP}{PB} = \frac{3}{5} = \frac{OA}{OB}\)

Thus, the verification holds.