(i) For \(ABC\) to be a straight line, vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) must be parallel. Calculate \(\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} 1 \\ p-5 \\ q+2 \end{pmatrix}\). Equating the ratios gives:

\(\frac{2}{1} = \frac{3}{p-5} = \frac{1}{q+2}\)

Solving these gives \(p = 6.5\) and \(q = -1.5\).

(ii) For \(\angle BAC = 90^\circ\), vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) must be perpendicular. Calculate \(\overrightarrow{AC} = \begin{pmatrix} 3 \\ p-2 \\ q+3 \end{pmatrix}\). The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = 0\) gives:

\(6 + 3(p-2) + (q+3) = 0\)

Solving gives \(q = -3p - 3\).

(iii) For \(p = 3\) and \(AB = AC\), equate the magnitudes:

\(AB^2 = 4 + 9 + 1 = 14\)

\(AC^2 = 9 + 1 + (q+3)^2\)

Equating gives \((q+3)^2 = 4\), so \(q+3 = \pm 2\).

Thus, \(q = -1\) or \(q = -5\).