(i) To show that the lines do not intersect, equate the components of the general points on l and m:

\(For l: r = 3i − j − 2k + λ(−i + j + 4k)\)

\(For m: r = 4i + 4j − 3k + μ(2i + j − 2k)\)

Equate at least two pairs of components and solve for λ or μ:

\(1. 3 − λ = 4 + 2μ\)

\(2. −1 + λ = 4 + μ\)

Solving these gives λ = \(\frac{3}{2}\) or μ = −\(\frac{7}{2}\).

Verify that not all three pairs of equations are satisfied, confirming the lines do not intersect.

(ii) To calculate the acute angle between the directions of the lines, find the direction vectors:

\(For l: d₁ = −i + j + 4k\)

\(For m: d₂ = 2i + j − 2k\)

\(Calculate the scalar product: d₁ ⋅ d₂ = (−1)(2) + (1)(1) + (4)(−2) = −2 + 1 − 8 = −9\)

Calculate the moduli: |d₁| = \(\sqrt{(-1)^2 + 1^2 + 4^2} = \sqrt{18}\)

|d₂| = \(\sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3\)

Divide the scalar product by the product of the moduli and evaluate the inverse cosine:

\(\cos \theta = \frac{-9}{\sqrt{18} \times 3} = \frac{-9}{9\sqrt{2}} = -\frac{1}{\sqrt{2}}\)

\(\theta = \cos^{-1}(-\frac{1}{\sqrt{2}}) = 45^{\circ}\) or \(\frac{1}{4} \pi\) (0.785 radians).

To find the vector equation of the line passing through points A and B, we use the position vectors:

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

\(\overrightarrow{OB} = 3\mathbf{i} + \mathbf{j} + \mathbf{k}\)

The direction vector \(\overrightarrow{AB}\) is given by:

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

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

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

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

Equating the components of the line \(l\) and the line through A and B:

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

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

\(m - 4\mu = 2 - \lambda\)

Solving these equations, we find \(\lambda = \frac{5}{7}\) and \(\mu = \frac{3}{7}\).

Substituting \(\mu = \frac{3}{7}\) into the third equation:

\(m - 4\left(\frac{3}{7}\right) = 2 - \frac{5}{7}\)

\(m - \frac{12}{7} = \frac{9}{7}\)

\(m = \frac{9}{7} + \frac{12}{7} = 3\)

Thus, the value of \(m\) is \(3\).

(a) To find the values of \(a\), \(b\), and \(c\), we use the fact that lines l and m are perpendicular. The direction vectors of l and m are \((c, -2, 4)\) and \((2, -3, 1)\) respectively. The scalar product of these vectors must be zero:

\(2c + 6 + 4 = 0\)

Solving gives \(c = -5\).

Since P lies on l, substitute \(\mathbf{r} = 4\mathbf{i} + 7\mathbf{j} - 2\mathbf{k}\) into the equation of l:

\(4 = a + \lambda c\)

\(7 = 3 - 2\lambda\)

\(-2 = b + 4\lambda\)

Solving these equations gives \(a = -6\), \(b = 6\), and \(\lambda = 2\).

(b) To find the position vector of R, first find the position vector of Q by setting the perpendicular from P to m:

\(\mathbf{OQ} = -\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)

Using the ratio \(PQ : QR = 2 : 3\), find \(\mathbf{OR}\):

\(\mathbf{OR} = \mathbf{OP} + \frac{5}{2} \mathbf{PQ}\)

\(\mathbf{OR} = \left(4, 7, -2\right) + \frac{5}{2} \left(-5, -2, 4\right)\)

\(\mathbf{OR} = \left(\frac{17}{2}, \frac{1}{2}, 8\right)\)

To find the foot of the perpendicular from A to the line l, consider a general point P on l with parameter \(\lambda\), given by:

\(\overrightarrow{OP} = (9 + 3\lambda)\mathbf{i} + (-1 - \lambda)\mathbf{j} + (8 + 2\lambda)\mathbf{k}\).

The vector \(\overrightarrow{AP}\) is:

\(\overrightarrow{AP} = (8 + 3\lambda)\mathbf{i} + (-3 - \lambda)\mathbf{j} + (4 + 2\lambda)\mathbf{k}\).

For \(\overrightarrow{AP}\) to be perpendicular to the direction vector of l, \(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\), their dot product must be zero:

\((8 + 3\lambda) \cdot 3 + (-3 - \lambda) \cdot (-1) + (4 + 2\lambda) \cdot 2 = 0\).

Simplifying, we get:

\(24 + 9\lambda + 3 + \lambda + 8 + 4\lambda = 0\).

\(14\lambda + 35 = 0\).

\(\lambda = -\frac{5}{2}\).

Substitute \(\lambda = -\frac{5}{2}\) back into \(\overrightarrow{OP}\) to find the foot of the perpendicular:

\(\overrightarrow{OP} = \left(9 + 3\left(-\frac{5}{2}\right)\right)\mathbf{i} + \left(-1 - \left(-\frac{5}{2}\right)\right)\mathbf{j} + \left(8 + 2\left(-\frac{5}{2}\right)\right)\mathbf{k}\).

\(\overrightarrow{OP} = \frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} + 3\mathbf{k}\).

The reflection of A in l is found by using the formula:

\(\overrightarrow{OR} = 2\overrightarrow{OP} - \overrightarrow{OA}\).

\(\overrightarrow{OR} = 2\left(\frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} + 3\mathbf{k}\right) - (\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})\).

\(\overrightarrow{OR} = (3\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})\).

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

Express the general point on the line l in component form:

\((1 + 2\lambda, 2 - \lambda, 1 + \lambda)\).

The distance from the origin is given by:

\(\sqrt{(1 + 2\lambda)^2 + (2 - \lambda)^2 + (1 + \lambda)^2} = \sqrt{10}\).

Squaring both sides, we have:

\((1 + 2\lambda)^2 + (2 - \lambda)^2 + (1 + \lambda)^2 = 10\).

Expanding and simplifying:

\(1 + 4\lambda + 4\lambda^2 + 4 - 4\lambda + \lambda^2 + 1 + 2\lambda + \lambda^2 = 10\).

Combine like terms:

\(6\lambda^2 + 2\lambda + 6 = 10\).

Rearrange to form a quadratic equation:

\(6\lambda^2 + 2\lambda - 4 = 0\).

Divide the entire equation by 2:

\(3\lambda^2 + \lambda - 2 = 0\).

Using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 1\), \(c = -2\):

\(\lambda = \frac{-1 \pm \sqrt{1 + 24}}{6} = \frac{-1 \pm 5}{6}\).

This gives \(\lambda = \frac{2}{3}\) or \(\lambda = -1\).

Substitute back to find the position vectors:

For \(\lambda = -1\):

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

For \(\lambda = \frac{2}{3}\):

\((1 + 2\left(\frac{2}{3}\right), 2 - \left(\frac{2}{3}\right), 1 + \left(\frac{2}{3}\right)) = \left(\frac{7}{3}, \frac{4}{3}, \frac{5}{3}\right)\).