(i) To find the cosine of angle BAC, we first find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

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

\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (3\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\)

The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) \cdot (2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) = 2 + 6 + 12 = 20\)

The magnitudes are \(|\overrightarrow{AB}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\) and \(|\overrightarrow{AC}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{49} = 7\)

Thus, \(\cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \frac{20}{3 \times 7} = \frac{20}{21}\)

(ii) To find the area of triangle ABC, we use the sine of angle BAC:

\(\sin \angle BAC = \sqrt{1 - \left(\frac{20}{21}\right)^2} = \sqrt{\frac{41}{441}} = \frac{\sqrt{41}}{21}\)

The area is \(\frac{1}{2} |\overrightarrow{AB}| |\overrightarrow{AC}| \sin \angle BAC = \frac{1}{2} \times 3 \times 7 \times \frac{\sqrt{41}}{21} = \frac{1}{2} \sqrt{41}\)