(i) The vector \(\overrightarrow{AB}\) is given by \(\overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\).

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

Thus, the unit vector in the direction of \(\overrightarrow{AB}\) is \(\pm \frac{1}{6} \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\).

(ii) For \(\angle AOC = 90^\circ\), the dot product \(\overrightarrow{OA} \cdot \overrightarrow{OC} = 0\).

\(\overrightarrow{OA} \cdot \overrightarrow{OC} = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 2 \\ p \end{pmatrix} = 4 + 6 - p = 0\).

Solving for \(p\), we get \(p = 10\).

(iii) The length of \(\overrightarrow{AD}\) is given by \(\sqrt{(-2)^2 + 3^2 + (q+1)^2} = 7\).

\((-2)^2 + 3^2 + (q+1)^2 = 49\).

\(4 + 9 + (q+1)^2 = 49\).

\((q+1)^2 = 36\).

Solving \((q+1)^2 = 36\) gives \(q+1 = 6\) or \(q+1 = -6\).

Thus, \(q = 5\) or \(q = -7\).