(a) The position vector of M is the midpoint of OA, which is \(\frac{1}{2}(6\mathbf{i} + 2\mathbf{j}) = 3\mathbf{i} + \mathbf{j}\).

The position vector of N is given by dividing the segment AB in the ratio 2:1. Thus, \(\overrightarrow{ON} = \frac{1}{3}(2\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{3}(2(6\mathbf{i} + 2\mathbf{j}) + (2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})) = \frac{10}{3}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

The direction vector \(\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \left( \frac{10}{3}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \right) - (3\mathbf{i} + \mathbf{j}) = \frac{1}{3}\mathbf{i} + \mathbf{j} + 2\mathbf{k}\).

Thus, the vector equation for the line through M and N is \(\mathbf{r} = 3\mathbf{i} + \mathbf{j} + \lambda \left( \frac{1}{3}\mathbf{i} + \mathbf{j} + 2\mathbf{k} \right)\).

(b) The line through O and B is \(\mathbf{r} = \mu(2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\).

Equating the components of \(\overrightarrow{MN}\) and \(\overrightarrow{OB}\), we solve for \(\lambda\) and \(\mu\):

\(\lambda = 3\) or \(\mu = 2\).

Thus, the position vector of P is \(4\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}\).

(c) To find angle OPM, calculate the scalar product of direction vectors for OP and MP:

\(\overrightarrow{OP} = 4\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}\)

\(\overrightarrow{MP} = (4\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}) - (3\mathbf{i} + \mathbf{j}) = \mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\)

Calculate the dot product and magnitudes:

\(\overrightarrow{OP} \cdot \overrightarrow{MP} = 4 \times 1 + 4 \times 3 + 6 \times 6 = 52\)

\(|\overrightarrow{OP}| = \sqrt{4^2 + 4^2 + 6^2} = \sqrt{68}\)

\(|\overrightarrow{MP}| = \sqrt{1^2 + 3^2 + 6^2} = \sqrt{46}\)

\(\cos \theta = \frac{52}{\sqrt{68} \times \sqrt{46}}\)

\(\theta = \cos^{-1}\left(\frac{52}{\sqrt{68} \times \sqrt{46}}\right) \approx 21.6^{\circ}\)