(i) To show that \(AB\) is perpendicular to \(BC\), calculate the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 9 \end{pmatrix}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\)

Calculate the dot product:

\(\overrightarrow{AB} \cdot \overrightarrow{BC} = 3 \times 2 + (-6) \times 1 + 9 \times 0 = 6 - 6 = 0\)

Since the dot product is zero, \(AB\) is perpendicular to \(BC\).

(ii) To show that \(ABCD\) is a trapezium, calculate \(\overrightarrow{DC}\):

\(\overrightarrow{DC} = \overrightarrow{OC} - \overrightarrow{OD} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 6 \end{pmatrix}\)

Check if \(\overrightarrow{AB}\) is parallel to \(\overrightarrow{DC}\):

\(\overrightarrow{AB} = k \overrightarrow{DC}\) where \(k = \frac{3}{2}\)

Since \(AB \parallel DC\), \(ABCD\) is a trapezium.

(iii) To find the area of \(ABCD\), calculate the magnitudes:

\(|\overrightarrow{AB}| = \sqrt{9 + 36 + 81} = \sqrt{126} = 11.22\)

\(|\overrightarrow{DC}| = \sqrt{4 + 16 + 36} = \sqrt{56} = 7.483\)

\(|\overrightarrow{BC}| = \sqrt{4 + 1 + 0} = \sqrt{5} = 2.236\)

Area = \(\frac{1}{2} (|\overrightarrow{AB}| + |\overrightarrow{DC}|) \times |\overrightarrow{BC}| = \frac{1}{2} (11.22 + 7.483) \times 2.236 = 20.92\)