(i) To find the angle \(AOB\), we use the dot product formula:

\(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\)

Given \(\mathbf{a} = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} -1 \\ 2 \\ 4 \end{pmatrix}\), the dot product is:

\(\mathbf{a} \cdot \mathbf{b} = (-3)(-1) + (6)(2) + (3)(4) = 3 + 12 + 12 = 27\)

The magnitudes are:

\(|\mathbf{a}| = \sqrt{(-3)^2 + 6^2 + 3^2} = \sqrt{54}\)

\(|\mathbf{b}| = \sqrt{(-1)^2 + 2^2 + 4^2} = \sqrt{21}\)

Thus, \(27 = \sqrt{54} \times \sqrt{21} \times \cos \theta\)

\(\cos \theta = \frac{27}{\sqrt{54} \times \sqrt{21}}\)

\(\theta = \cos^{-1}\left(\frac{27}{\sqrt{54} \times \sqrt{21}}\right) \approx 36.7^\circ\) or \(0.641\) radians

(ii) To find \(\overrightarrow{OC}\), first find \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}\).

Then, \(\overrightarrow{AC} = 3\overrightarrow{AB} = 3 \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ -12 \\ 3 \end{pmatrix}\).

\(\overrightarrow{OC} = \mathbf{a} + \overrightarrow{AC} = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -12 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix}\).

The magnitude of \(\overrightarrow{OC}\) is \(\sqrt{3^2 + (-6)^2 + 6^2} = 9\).

The unit vector in the direction of \(\overrightarrow{OC}\) is \(\frac{1}{9} \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3} \end{pmatrix}\).

Thus, the unit vector is \(\frac{1}{9} \mathbf{i} - \frac{2}{3} \mathbf{j} + \frac{2}{3} \mathbf{k}\).