(i) To find the values of \(p\) for which angle \(AOB\) is 90°, we need \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\).

Calculate the dot product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (3p)(-p) + (4)(-1) + (p^2)(p^2) = -3p^2 - 4 + p^4\).

Set the dot product to zero:

\(-3p^2 - 4 + p^4 = 0\).

Rearrange to form a quadratic in \(p^2\):

\(p^4 - 3p^2 - 4 = 0\).

Factorize:

\((p^2 + 1)(p^2 - 4) = 0\).

Thus, \(p^2 = -1\) (no real solution) or \(p^2 = 4\).

Therefore, \(p = \pm 2\).

(ii) For \(p = 3\), find \(\overrightarrow{BA}\):

\(\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = \begin{pmatrix} 3(3) \\ 4 \\ 3^2 \end{pmatrix} - \begin{pmatrix} -3 \\ -1 \\ 9 \end{pmatrix} = \begin{pmatrix} 9 \\ 4 \\ 9 \end{pmatrix} - \begin{pmatrix} -3 \\ -1 \\ 9 \end{pmatrix} = \begin{pmatrix} 12 \\ 5 \\ 0 \end{pmatrix}\).

Calculate the magnitude of \(\overrightarrow{BA}\):

\(|\overrightarrow{BA}| = \sqrt{12^2 + 5^2 + 0^2} = \sqrt{144 + 25} = \sqrt{169} = 13\).

The unit vector in the direction of \(\overrightarrow{BA}\) is:

\(\frac{1}{13} \begin{pmatrix} 12 \\ 5 \\ 0 \end{pmatrix}\).