(a) To show that angle ABC is a right angle, we need to find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\).

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) - (2\mathbf{i} + 5\mathbf{k}) = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\mathbf{i} + \mathbf{j} + \mathbf{k}) - (3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = -2\mathbf{i} - \mathbf{j} - 2\mathbf{k}\)

Calculate the scalar product: \(\overrightarrow{AB} \cdot \overrightarrow{BC} = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) \cdot (-2\mathbf{i} - \mathbf{j} - 2\mathbf{k}) = -2 - 2 + 4 = 0\)

Since the scalar product is zero, angle ABC is a right angle.

(b) To show that triangle ABC is isosceles, calculate the lengths of \(AB\) and \(BC\).

\(|\overrightarrow{AB}| = \sqrt{(1)^2 + (2)^2 + (-2)^2} = \sqrt{9} = 3\)

\(|\overrightarrow{BC}| = \sqrt{(-2)^2 + (-1)^2 + (-2)^2} = \sqrt{9} = 3\)

Since \(AB = BC\), triangle ABC is isosceles.

(c) To find the exact length of the perpendicular from O to the line through B and C, use the vector equation of the line: \(\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(-2\mathbf{i} - \mathbf{j} - 2\mathbf{k})\).

Taking a general point \(P\) on the line, form an equation by setting the derivative of \(|\overrightarrow{OP}|\) to zero or using the scalar product method.

Solve and obtain \(\lambda = -\frac{5}{9}\).

Calculate the perpendicular distance: \(\frac{1}{3} \sqrt{2}\).