(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 \(5\mathbf{i} - 8\mathbf{j} + 7\mathbf{k}\).

(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.

(i) The vector equation of the line l passing through A and parallel to OB is given by:

\(\mathbf{r} = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}\)

where \(\lambda\) is a scalar.

(ii) To find the position vector of N, express \(\overrightarrow{BN}\) in terms of \(\lambda\):

\(\overrightarrow{BN} = \begin{pmatrix} -1 + 3\lambda \\ 3 - \lambda \\ 5 - 4\lambda \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}\)

Equate its scalar product with \(\begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}\) to zero and solve for \(\lambda\):

\((3)(-1 + 3\lambda) + (-1)(3 - \lambda) + (-4)(5 - 4\lambda) = 0\)

Simplifying gives \(\lambda = 2\).

Substitute \(\lambda = 2\) back to find \(\overrightarrow{ON}\):

\(\overrightarrow{ON} = \begin{pmatrix} -1 + 6 \\ 3 - 2 \\ 5 - 8 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \\ -3 \end{pmatrix}\)

To show \(BN = 3\), calculate the magnitude of \(\overrightarrow{BN}\):

\(\overrightarrow{BN} = \begin{pmatrix} 5 - 3 \\ 1 + 1 \\ -3 + 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}\)

\(BN = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\)

First, find the direction vector for the line through A and B:

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

The equation of the line through A and B is:

\(\mathbf{r} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + t(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\).

Equate the vector equations of the lines to find if they intersect:

\(4\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} + s(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + t(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\).

Equating components, we get:

1. \(4 + s = 2 - t\)

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

3. \(2 + s = 1 + 2t\)

Solving these equations, we find:

From equation 1: \(s + t = -2\)

From equation 2: \(s - t = 2\)

Adding these, \(2s = 0 \Rightarrow s = 0\).

Substituting \(s = 0\) into \(s + t = -2\), we get \(t = -2\).

Substitute \(s = 0\) and \(t = -2\) into equation 3:

\(2 + 0 \neq 1 + 2(-2) \Rightarrow 2 \neq -3\).

Since the equations are inconsistent, the lines do not intersect.

(a) To find the obtuse angle between \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\), calculate the scalar product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (3)(1) + (-1)(2) + (2)(-3) = 3 - 2 - 6 = -5\)

Calculate the magnitudes:

\(|\overrightarrow{OA}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{14}\)

\(|\overrightarrow{OB}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14}\)

Use the cosine formula:

\(\cos \theta = \frac{-5}{\sqrt{14} \times \sqrt{14}} = \frac{-5}{14}\)

\(\theta = \cos^{-1}\left(\frac{-5}{14}\right) \approx 110.9^\circ \text{ or } 1.94 \text{ radians}\)

(b) The vector equation for the line \(l\) through points \(A\) and \(B\) is:

\(\mathbf{r} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + \mu_1 \begin{pmatrix} 1 - 3 \\ 2 + 1 \\ -3 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + \mu_1 \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}\)

(c) For the line through \(C\) and \(D\):

\(\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 5 - 1 \\ -6 + 2 \\ 11 - 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -4 \\ 6 \end{pmatrix}\)

Equate components and solve for \(\lambda\) and \(\mu_1\):

\(3 + 2\mu_1 = 1 + 4\lambda\)

\(-1 - 3\mu_1 = -2 - 4\lambda\)

\(2 + 5\mu_1 = 5 + 6\lambda\)

Solve to find \(\mu_1 = 3\) and \(\lambda = 2\).

Substitute back to find the position vector:

\(\mathbf{r} = \begin{pmatrix} 9 \\ -10 \\ 17 \end{pmatrix}\)