(i) To find angle \(BAC\), we need vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 6 \\ -1 \\ 7 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix}\)

\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}\)

The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = 4 \times 0 + (-2) \times 3 + 4 \times 4 = 0 - 6 + 16 = 10\)

The magnitudes are \(|\overrightarrow{AB}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{36} = 6\)

\(|\overrightarrow{AC}| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{25} = 5\)

Using the cosine formula: \(\cos BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| \times |\overrightarrow{AC}|} = \frac{10}{6 \times 5} = \frac{1}{3}\)

Thus, \(BAC = \cos^{-1}\left(\frac{1}{3}\right)\).

(ii) To find the area of triangle \(ABC\), use \(\sin BAC = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3}\)

Area = \(\frac{1}{2} \times |\overrightarrow{AB}| \times |\overrightarrow{AC}| \times \sin BAC = \frac{1}{2} \times 6 \times 5 \times \frac{\sqrt{8}}{3} = 5\sqrt{8}\)