(i) The vector equation of the line through A and B is \(\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(\mathbf{i} - \mathbf{j} + 2\mathbf{k})\).

Equating components with the line \(l\):

\(1 + \lambda = 2 - \mu\)

\(1 - \lambda = -2 + 2\mu\)

\(1 + 2\lambda = -1 + \mu\)

Solving these equations, we find \(\lambda = -1\) and \(\mu = 2\), but the third equation is not satisfied, showing no intersection.

(ii) To find the perpendicular distance from A to \(l\), consider a point \(P\) on \(l\) given by \((1-\mu)\mathbf{i} + (-3 + 2\mu)\mathbf{j} + (-2 + \mu)\mathbf{k}\).

Calculate \(\overrightarrow{AP}\) and find \(\mu\) such that \(\overrightarrow{AP}\) is perpendicular to the direction vector of \(l\), \(-\mathbf{i} + 2\mathbf{j} + \mathbf{k}\).

Solving gives \(\mu = \frac{3}{2}\).

Substitute \(\mu = \frac{3}{2}\) into \(\overrightarrow{AP}\) and find its magnitude: \(\frac{1}{\sqrt{2}}\).

To find the position vector of D, we use the property of a parallelogram that opposite sides are equal and parallel. Therefore, \(\overrightarrow{AB} = \overrightarrow{CD}\).

First, calculate \(\overrightarrow{AB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (0\mathbf{i} + 4\mathbf{j} + \mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = -\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\).

Since \(\overrightarrow{CD} = \overrightarrow{AB}\), we have:

\(\overrightarrow{CD} = -\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\).

Now, find \(\overrightarrow{OD}\):

\(\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{CD} = (2\mathbf{i} + 5\mathbf{j} - \mathbf{k}) + (-\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = 3\mathbf{i} + 3\mathbf{j} + \mathbf{k}\).

To verify that ABCD is a rhombus, we need to show that all sides are equal in length. Calculate the magnitudes of \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\):

\(|\overrightarrow{AB}| = \sqrt{(-1)^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\).

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (2\mathbf{i} + 5\mathbf{j} - \mathbf{k}) - (0\mathbf{i} + 4\mathbf{j} + \mathbf{k}) = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\).

\(|\overrightarrow{BC}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3\).

Since \(|\overrightarrow{AB}| = |\overrightarrow{BC}|\), ABCD has a pair of adjacent sides that are equal, confirming it is a rhombus.

First, find the vector equation of the line passing through A and B. The direction vector \(\overrightarrow{AB}\) is given by:

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (\mathbf{i} + \mathbf{j} + 5\mathbf{k}) - (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\)

The vector equation of the line through A and B is:

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

Equate the components of the general points on both lines:

\(2 - \lambda = 1 + 3\mu\)

\(-1 + 2\lambda = 1 + \mu\)

\(3 + 2\lambda = 2 - \mu\)

Solve these equations for \(\lambda\) and \(\mu\):

From the first equation: \(\lambda = 1 - 3\mu\)

Substitute into the second equation:

\(-1 + 2(1 - 3\mu) = 1 + \mu\)

\(-1 + 2 - 6\mu = 1 + \mu\)

\(-6\mu - \mu = 1\)

\(-7\mu = 0\)

\(\mu = 0\)

Substitute \(\mu = 0\) into \(\lambda = 1 - 3\mu\):

\(\lambda = 1\)

Check the third equation:

\(3 + 2(1) = 2 - 0\)

\(5 \neq 2\)

Since not all three equations are satisfied, the lines do not intersect.

(i) The direction vector of \(l_1\) is obtained from the points \((0, 1, 5)\) and \((2, -2, 1)\), which is \(\begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix}\).

The direction vector of \(l_2\) is \(\begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}\).

Since the direction vectors are not parallel, the lines are not parallel.

Express the general point of \(l_1\) as \((2\lambda, 1 - 3\lambda, 5 - 4\lambda)\) and \(l_2\) as \((7 + \mu, 1 + 2\mu, 1 + 5\mu)\).

Equate at least two pairs of components and solve for \(\lambda\) and \(\mu\). Verify that all three component equations are not satisfied, confirming the lines are skew.

(ii) The direction vector of \(l_2\) is \(\begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}\) and the direction of the \(x\)-axis is \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).

The scalar product is \(1 \times 1 + 2 \times 0 + 5 \times 0 = 1\).

The magnitudes are \(\sqrt{1^2 + 2^2 + 5^2} = \sqrt{30}\) and \(\sqrt{1^2} = 1\).

The cosine of the angle is \(\frac{1}{\sqrt{30}}\).

The angle is \(\cos^{-1}\left(\frac{1}{\sqrt{30}}\right) \approx 79.5^\circ\) or \(1.39\) radians.

(i) To show that the lines intersect, equate the vector equations:

\(1 + \lambda = a + \mu\)

\(4 = 2 + 2\mu\)

\(-2 + 3\lambda = -2 + 3a\mu\)

From the second equation, solve for \(\mu\):

\(2 = 2\mu \Rightarrow \mu = 1\)

Substitute \(\mu = 1\) into the first equation:

\(1 + \lambda = a + 1 \Rightarrow \lambda = a\)

Check the third equation:

\(-2 + 3a = -2 + 3a\)

All equations are satisfied, so the lines intersect for all \(a\).

(ii) The point of intersection is \((a + 1, 4, 3a - 2)\).

The distance from the origin is given by:

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

\((a+1)^2 + 16 + (3a-2)^2 = 81\)

\(a^2 + 2a + 1 + 16 + 9a^2 - 12a + 4 = 81\)

\(10a^2 - 10a + 21 = 81\)

\(10a^2 - 10a - 60 = 0\)

\(a^2 - a - 6 = 0\)

Factor the quadratic:

\((a - 3)(a + 2) = 0\)

Thus, \(a = 3\) or \(a = -2\).