(i) To find angle \(ABC\), we use the vectors \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\).

\(\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 10 \\ 0 \\ 6 \end{pmatrix} - \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}\)

The dot product \(\overrightarrow{BA} \cdot \overrightarrow{BC} = (-2)(6) + (1)(-2) + (2)(3) = -8\)

The magnitudes are \(|\overrightarrow{BA}| = \sqrt{(-2)^2 + 1^2 + 2^2} = 3\) and \(|\overrightarrow{BC}| = \sqrt{6^2 + (-2)^2 + 3^2} = 7\).

Using the dot product formula: \(\overrightarrow{BA} \cdot \overrightarrow{BC} = |\overrightarrow{BA}| |\overrightarrow{BC}| \cos \theta\)

\(-8 = 3 \times 7 \times \cos \theta\)

\(\cos \theta = \frac{-8}{21}\)

\(\theta = \cos^{-1}\left(\frac{-8}{21}\right) \approx 112.4^\circ\) or \(1.96\) radians.

(ii) Since ABCD is a parallelogram, \(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{BC}\).

\(\overrightarrow{OD} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix} + \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \\ 8 \end{pmatrix}\).