1. Find \(\overrightarrow{PQ}\) for a general point Q on l, e.g., \(-3\mathbf{i} + 6\mathbf{k} + \mu(2\mathbf{i} - \mathbf{j} - 2\mathbf{k})\).

2. Calculate the scalar product of \(\overrightarrow{PQ}\) and a direction vector for l and equate the result to zero.

3. Solve for \(\mu\) and obtain \(\mu = 2\).

4. Carry out a complete method for finding the length of \(\overrightarrow{PQ}\).

5. Obtain answer 3.

First, find the vector equation of the line passing through A and B. The direction vector AB is given by:

\(\text{Direction vector } \overrightarrow{AB} = (3\mathbf{i} + \mathbf{j} + \mathbf{k}) - (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\)

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

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

Equate the components of the general points on both lines:

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

Solving these equations:

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

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

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

From equation 1: \(\lambda = \frac{1}{2} \mu + \frac{1}{2}\)

From equation 2: \(\lambda = -\mu + 1\)

From equation 3: \(\lambda = \mu + 1\)

Substitute \(\lambda = \mu + 1\) into \(\lambda = \frac{1}{2} \mu + \frac{1}{2}\):

\(\mu + 1 = \frac{1}{2} \mu + \frac{1}{2}\)

\(\mu = -1\)

Substitute \(\mu = -1\) into \(\lambda = \mu + 1\):

\(\lambda = 0\)

Verify with the third equation:

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

\(-1 \neq -1\)

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

(i) The vector equation of the line passing through A and B is \(\mathbf{r} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k} + \lambda(2\mathbf{i} - 2\mathbf{k})\).

Equating the components of the line l and the line through A and B gives:

\(4 + \mu = 2 + 2\lambda\)

\(6 + 2\mu = 1\)

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

Solving these equations, we find that they are inconsistent, confirming that the lines do not intersect.

(ii) The direction vector for \(\overrightarrow{AP}\) is \((2 + t, 5 + 2t, -3 - 2t)\).

Using the cosine of the angle formula, \(\cos 120^{\circ} = -\frac{1}{2}\), we set up the equation:

\(\frac{(2 + t)(2) + (5 + 2t)(0) + (-3 - 2t)(-2)}{\sqrt{(2 + t)^2 + (5 + 2t)^2 + (-3 - 2t)^2} \cdot \sqrt{4 + 0 + 4}} = -\frac{1}{2}\)

Simplifying, we obtain \(3t^2 + 8t + 4 = 0\).

Solving this quadratic equation, we find \(t = -2\) or \(t = -\frac{2}{3}\). Using \(t = -2\), the position vector of P is \(2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\).

To determine if the lines are skew, we need to check if they are neither parallel nor intersecting.

First, equate the components of the lines:

For the i component: \(2 + 2s = 1 + t\)

For the j component: \(-1 + 3s = 3 + 2t\)

For the k component: \(1 - s = 4 + t\)

Solving these equations:

From \(2 + 2s = 1 + t\), we get \(t = 2s - 1\).

From \(-1 + 3s = 3 + 2t\), substitute \(t = 2s - 1\):

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

\(-1 + 3s = 3 + 4s - 2\)

\(-1 + 3s = 1 + 4s\)

\(-2 = s\)

Substitute \(s = -2\) into \(t = 2s - 1\):

\(t = 2(-2) - 1 = -4 - 1 = -5\)

Check the k component: \(1 - s = 4 + t\)

\(1 - (-2) = 4 + (-5)\)

\(3 \neq -1\)

The equations do not satisfy all components simultaneously, so the lines do not intersect.

Since the direction vectors \((2, 3, -1)\) and \((1, 2, 1)\) are not scalar multiples, the lines are not parallel.

Therefore, the lines are skew.

To find the length of the perpendicular from point \(P\) to the line \(l\), we can use the vector projection method.

1. The direction vector of the line \(l\) is \(\mathbf{d} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\).

2. The position vector of point \(P\) is \(\mathbf{p} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\).

3. A point \(A\) on the line \(l\) can be represented as \(\mathbf{a} = 4\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}\).

4. The vector \(\mathbf{PA}\) is \(\mathbf{p} - \mathbf{a} = (3 - 4)\mathbf{i} + (-2 - 2)\mathbf{j} + (1 - 5)\mathbf{k} = -\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}\).

5. The projection of \(\mathbf{PA}\) onto \(\mathbf{d}\) is given by:

\(\text{proj}_{\mathbf{d}} \mathbf{PA} = \frac{\mathbf{PA} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d}\)

6. Calculate \(\mathbf{PA} \cdot \mathbf{d} = (-1)(1) + (-4)(2) + (-4)(3) = -1 - 8 - 12 = -21\).

7. Calculate \(\mathbf{d} \cdot \mathbf{d} = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14\).

8. The projection is:

\(\text{proj}_{\mathbf{d}} \mathbf{PA} = \frac{-21}{14} \mathbf{d} = -\frac{3}{2} \mathbf{d}\)

9. The perpendicular vector \(\mathbf{v}\) from \(P\) to \(l\) is:

\(\mathbf{v} = \mathbf{PA} - \text{proj}_{\mathbf{d}} \mathbf{PA}\)

10. Calculate \(\mathbf{v} = (-\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}) - \left(-\frac{3}{2}(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\right)\).

11. Simplify \(\mathbf{v} = -\mathbf{i} - 4\mathbf{j} - 4\mathbf{k} + \frac{3}{2}\mathbf{i} + 3\mathbf{j} + \frac{9}{2}\mathbf{k}\).

12. \(\mathbf{v} = \left(-1 + \frac{3}{2}\right)\mathbf{i} + (-4 + 3)\mathbf{j} + \left(-4 + \frac{9}{2}\right)\mathbf{k}\).

13. \(\mathbf{v} = \frac{1}{2}\mathbf{i} - \mathbf{j} + \frac{1}{2}\mathbf{k}\).

14. The length of \(\mathbf{v}\) is:

\(\|\mathbf{v}\| = \sqrt{\left(\frac{1}{2}\right)^2 + (-1)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + 1 + \frac{1}{4}} = \sqrt{\frac{3}{2}}\)

15. \(\|\mathbf{v}\| \approx 1.22\) to 3 significant figures.