(i) For OAB to be a straight line, \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) must be parallel. Thus, \(\begin{pmatrix} 4 \\ 2 \\ p \end{pmatrix} = k \begin{pmatrix} p \\ 1 \\ 1 \end{pmatrix}\) for some scalar \(k\). Solving, we find \(p = 2\). The unit vector in the direction of \(\overrightarrow{OA}\) is \(\frac{1}{\sqrt{6}} \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\).

(ii) For OA to be perpendicular to AB, \(\overrightarrow{OA} \cdot \overrightarrow{AB} = 0\). First, find \(\overrightarrow{AB} = \begin{pmatrix} 4-p \\ 1 \\ p-1 \end{pmatrix}\). Then, \(\overrightarrow{OA} \cdot \overrightarrow{AB} = (p)(4-p) + 1 + (p-1) = 0\). Simplifying gives \(5p - p^2 = 0\), so \(p = 0\) or \(p = 5\).

(iii) For OABC to be a parallelogram, \(\overrightarrow{OC} = \overrightarrow{OB} - \overrightarrow{OA}\). With \(p = 3\), \(\overrightarrow{OB} = \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix}\) and \(\overrightarrow{OA} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\). Thus, \(\overrightarrow{OC} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\).