(i) The position vector of the midpoint \(M\) of \(AB\) is:

\(\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{2}((\mathbf{i} - \mathbf{k}) + (3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k})) = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\)

The point \(N\) divides \(AC\) in the ratio \(2:1\), so:

\(\overrightarrow{ON} = \frac{2\overrightarrow{OC} + \overrightarrow{OA}}{3} = \frac{2(4\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) + (\mathbf{i} - \mathbf{k})}{3} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\)

The direction vector of \(MN\) is:

\(\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = (3\mathbf{i} - 2\mathbf{j} + \mathbf{k}) - (2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = \mathbf{i} - 3\mathbf{j} + 3\mathbf{k}\)

Thus, the vector equation of the line \(MN\) is:

\(\mathbf{r} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 3\mathbf{j} + 3\mathbf{k})\)

(ii) The equation of line \(BC\) is:

\(\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} + \mu(\mathbf{i} - 5\mathbf{j} + 5\mathbf{k})\)

Setting the equations of \(MN\) and \(BC\) equal gives:

\(2\mathbf{i} + \mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}) = 3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} + \mu(\mathbf{i} - 5\mathbf{j} + 5\mathbf{k})\)

Equating components, we solve for \(\lambda\) and \(\mu\):

\(\lambda = 3, \quad \mu = 2\)

Substituting \(\mu = 2\) into the equation of \(BC\):

\(\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} + 2(\mathbf{i} - 5\mathbf{j} + 5\mathbf{k}) = 5\mathbf{i} - 8\mathbf{j} + 7\mathbf{k}\)

Thus, the position vector of \(P\) is \(5\mathbf{i} - 8\mathbf{j} + 7\mathbf{k}\).