(i) To find \(\overrightarrow{OB}\), use \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = (\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) + (3\mathbf{i} - \mathbf{j} + \mathbf{k}) = 4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\).

The magnitude of \(\overrightarrow{OB}\) is \(\sqrt{4^2 + 2^2 + 4^2} = 6\).

The unit vector in the direction of \(\overrightarrow{OB}\) is \(\frac{1}{6}(4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})\).

(ii) To find \(\overrightarrow{AC}\), use \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (3\mathbf{i} - \mathbf{j} + \mathbf{k}) - (\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}\).

The magnitudes are \(|\overrightarrow{OB}| = 6\) and \(|\overrightarrow{AC}| = \sqrt{24}\).

The dot product \(\overrightarrow{AC} \cdot \overrightarrow{OB} = 8 - 8 - 8 = -8\).

Using \(\overrightarrow{AC} \cdot \overrightarrow{OB} = |\overrightarrow{AC}| |\overrightarrow{OB}| \cos \theta\), we have \(-8 = 6 \times \sqrt{24} \times \cos \theta\).

Solving for \(\theta\), \(\theta = 105.8^\circ\), so the acute angle is \(74.2^\circ\).

(iii) The magnitudes are \(|\overrightarrow{OA}| = \sqrt{19}\) and \(|\overrightarrow{OC}| = \sqrt{11}\).

The perimeter is \(2(\sqrt{19} + \sqrt{11}) \approx 15.4\).