(i) To find the value of \(p\) when \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow{OB}\), set the scalar product to zero:

\((p-6)(4-2p) + (2p-6)p + 1 \times 2 = 0\)

\(-2p^2 + 16p - 24 + 2p^2 - 6p + 2 = 0\)

\(10p - 22 = 0\)

Solving for \(p\), we get \(p = 2.2\).

(ii) For \(OAB\) to be a straight line, \(\overrightarrow{OB} = 2 \overrightarrow{OA}\):

\(4-2p = 2(p-6)\) and \(p = 2(2p-6)\)

Solving \(4-2p = 2(p-6)\), we get \(p = 4\).

Thus, \(\overrightarrow{OA} = \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} -4 \\ 4 \\ 2 \end{pmatrix}\).

The length of \(OA\) is \(\sqrt{(-2)^2 + 2^2 + 1^2} = 3\).