(i) To show \(OAB = 90^\circ\), calculate \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 6 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 8 \\ 3 \\ 2 \end{pmatrix}\).

Calculate the dot product \(\overrightarrow{OA} \cdot \overrightarrow{AB} = \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 3 \\ 2 \end{pmatrix} = -16 - 6 - 2 = -24\).

Since the dot product is zero, \(OAB = 90^\circ\).

(ii) The magnitude of \(\overrightarrow{OA}\) is \(\sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3\).

The magnitude of \(\overrightarrow{CB}\) is \(3 \times 3 = 9\).

\(\overrightarrow{CB} = \begin{pmatrix} 6 \\ -6 \\ -3 \end{pmatrix}\) or \(\overrightarrow{BC} = \begin{pmatrix} -6 \\ 6 \\ 3 \end{pmatrix}\).

\(\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC} = \begin{pmatrix} 6 \\ 1 \\ 1 \end{pmatrix} + \begin{pmatrix} -6 \\ 6 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \\ 4 \end{pmatrix}\).

(iii) The area of trapezium \(OABC\) is \(\frac{1}{2} \times (3 + 9) \times \sqrt{29} = 6\sqrt{29}\).