(i) The vector equation for the line through A and B is \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + s(\mathbf{i} - \mathbf{j})\). Equating components with the line \(l\), we have:

\(1 + s = 1 - 2t\)

\(2 - s = 5 + t\)

\(3 = 2 - t\)

Solving these gives \(s = -6, t = 3\) for the first equation, \(s = 2, t = -1\) for the second, and \(t = -1\) for the third. The values do not satisfy all equations simultaneously, so \(l\) does not intersect the line through A and B.

(ii) The direction vector for \(AP\) is \((-2t, 3 + t, -1 - t)\). The cosine of angle \(PAB\) is given by:

\(\cos 60^{\circ} = \frac{\overrightarrow{AP} \cdot \overrightarrow{AB}}{|\overrightarrow{AP}||\overrightarrow{AB}|}\)

Expanding the scalar product and expressing the product of the moduli in terms of \(t\), we obtain the equation \(3t^2 + 7t + 2 = 0\).

Solving the quadratic equation, we find roots \(t = -2\) and \(t = -\frac{1}{3}\). The position vector of P for \(t = -2\) is \(5\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\), which is valid as it satisfies the angle condition.