(i) To find the value of \(k\) for which angle \(AOB\) is \(90^\circ\), we use the dot product formula:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\).

\(\begin{pmatrix} 6 \\ -2 \\ -6 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ k \\ -3 \end{pmatrix} = 6 \times 3 + (-2) \times k + (-6) \times (-3) = 0\).

\(18 - 2k + 18 = 0\).

\(36 - 2k = 0\).

\(k = 18\).

(ii) To find the values of \(k\) for which the lengths of \(OA\) and \(OB\) are equal, we equate their magnitudes:

\(\sqrt{6^2 + (-2)^2 + (-6)^2} = \sqrt{3^2 + k^2 + (-3)^2}\).

\(\sqrt{36 + 4 + 36} = \sqrt{9 + k^2 + 9}\).

\(\sqrt{76} = \sqrt{k^2 + 18}\).

\(76 = k^2 + 18\).

\(k^2 = 58\).

\(k = \pm \sqrt{58}\) or \(k \approx \pm 7.62\).

(iii) For \(k = 4\), find the unit vector in the direction of \(\overrightarrow{OC}\).

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

\(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix} 6 \\ -2 \\ -6 \end{pmatrix} + \begin{pmatrix} -2 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ -4 \end{pmatrix}\).

The unit vector is \(\frac{1}{6} \begin{pmatrix} 4 \\ 2 \\ -4 \end{pmatrix}\) or \(\frac{1}{6} (4\mathbf{i} + 2\mathbf{j} - 4\mathbf{k})\).