(i) To show that \(\angle ABC = 90^\circ\), we need to show that vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are perpendicular.

First, find \(\overrightarrow{AB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} - \begin{pmatrix} 3 \\ 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}.\)

Next, find \(\overrightarrow{BC}\):

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 6 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}.\)

Calculate the dot product \(\overrightarrow{AB} \cdot \overrightarrow{BC}\):

\(\overrightarrow{AB} \cdot \overrightarrow{BC} = 2 \times 1 + (-3) \times 2 + 1 \times 4 = 2 - 6 + 4 = 0.\)

Since the dot product is zero, \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are perpendicular, so \(\angle ABC = 90^\circ\).

(ii) To find the area of triangle \(ABC\), use the formula:

\(\text{Area} = \frac{1}{2} \times |\overrightarrow{AB}| \times |\overrightarrow{BC}|.\)

Calculate the magnitudes:

\(|\overrightarrow{AB}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}.\)

\(|\overrightarrow{BC}| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{21}.\)

Thus, the area is:

\(\text{Area} = \frac{1}{2} \times \sqrt{14} \times \sqrt{21} = \frac{1}{2} \times \sqrt{294}.\)

\(\sqrt{294} \approx 17.146,\) so the area is approximately \(8.6\) to 1 decimal place.