(a) To show \(OA = OB\), calculate the magnitudes:

\(|\overrightarrow{OA}| = \sqrt{(2)^2 + (-1)^2 + (0)^2} = \sqrt{4 + 1} = \sqrt{5}\)

\(|\overrightarrow{OB}| = \sqrt{(0)^2 + (1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}\)

Thus, \(OA = OB = \sqrt{5}\).

To find angle \(AOB\), use the scalar product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (2)(0) + (-1)(1) + (0)(-2) = -1\)

\(\cos \theta = \frac{-1}{\sqrt{5} \times \sqrt{5}} = \frac{-1}{5}\)

\(\theta = \cos^{-1}\left(\frac{-1}{5}\right) \approx 101.5^{\circ}\)

(b) The midpoint \(M\) of \(AB\) is:

\(M = \left(\frac{2+0}{2}, \frac{-1+1}{2}, \frac{0-2}{2}\right) = (1, 0, -1)\)

Position vector of \(M\) is \(\mathbf{i} - \mathbf{k}\).

Let \(\overrightarrow{OP} = \lambda(\mathbf{i} - \mathbf{k})\).

Given \(PA : OA = \sqrt{7} : 1\), express \(AP = \sqrt{7} \cdot OA\):

\(\overrightarrow{AP} = \lambda(\mathbf{i} - \mathbf{k}) - (2\mathbf{i} - \mathbf{j})\)

\(\sqrt{(\lambda - 2)^2 + (-\lambda + 1)^2 + (-\lambda)^2} = \sqrt{7} \cdot \sqrt{5}\)

\((\lambda - 2)^2 + (-\lambda + 1)^2 + (-\lambda)^2 = 35\)

Solving gives \(\lambda = 5\) or \(\lambda = -3\).

Thus, \(\overrightarrow{OP} = 5(\mathbf{i} - \mathbf{k}) = 5\mathbf{i} - 5\mathbf{k}\) or \(-3(\mathbf{i} - \mathbf{k}) = -3\mathbf{i} + 3\mathbf{k}\).