To find the length of the perpendicular from point A to the line l, we first need to find the vector AP where P is a point on the line l.

The line l is given by r = 4i - 9j + 9k + \(\lambda (-2i + j - 2k)\).

Let P be a point on the line, then the position vector of P is r = (4 - 2\(\lambda\))i + (-9 + \(\lambda\))j + (9 - 2\(\lambda\))k.

The position vector of A is 3i + 8j + 5k.

Thus, AP = ((4 - 2\(\lambda\)) - 3)i + ((-9 + \(\lambda\)) - 8)j + ((9 - 2\(\lambda\)) - 5)k.

Simplifying, AP = (1 - 2\(\lambda\))i + (-17 + \(\lambda\))j + (4 - 2\(\lambda\))k.

To find the perpendicular distance, we need to minimize the magnitude of AP.

The magnitude of AP is \(\sqrt{(1 - 2\lambda)^2 + (-17 + \lambda)^2 + (4 - 2\lambda)^2}\).

Expanding and simplifying, we get:

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

\((-17 + \lambda)^2 = 289 - 34\lambda + \lambda^2\)

\((4 - 2\lambda)^2 = 16 - 16\lambda + 4\lambda^2\)

Adding these, we get:

\(5\lambda^2 - 54\lambda + 306\)

To minimize, set the derivative with respect to \(\lambda\) to zero:

\(10\lambda - 54 = 0\)

\(\lambda = 5.4\)

Substitute \(\lambda = 5.4\) back into the expression for the magnitude:

\(\sqrt{5(5.4)^2 - 54(5.4) + 306} = 15\)

Thus, the length of the perpendicular from A to l is 15.

(i) To find the cosine of angle BAC, we first find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

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

\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (3\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\)

The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) \cdot (2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) = 2 + 6 + 12 = 20\)

The magnitudes are \(|\overrightarrow{AB}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\) and \(|\overrightarrow{AC}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{49} = 7\)

Thus, \(\cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \frac{20}{3 \times 7} = \frac{20}{21}\)

(ii) To find the area of triangle ABC, we use the sine of angle BAC:

\(\sin \angle BAC = \sqrt{1 - \left(\frac{20}{21}\right)^2} = \sqrt{\frac{41}{441}} = \frac{\sqrt{41}}{21}\)

The area is \(\frac{1}{2} |\overrightarrow{AB}| |\overrightarrow{AC}| \sin \angle BAC = \frac{1}{2} \times 3 \times 7 \times \frac{\sqrt{41}}{21} = \frac{1}{2} \sqrt{41}\)

To find the intersection, equate the parametric equations of the lines:

\(\begin{pmatrix} 5 + s \\ 1 - s \\ -4 + 3s \end{pmatrix} = \begin{pmatrix} p + 2t \\ 4 + 5t \\ -2 - 4t \end{pmatrix}\).

Equating components, we get:

1. \(5 + s = p + 2t\)

2. \(1 - s = 4 + 5t\)

3. \(-4 + 3s = -2 - 4t\)

From equation 2: \(s = -3 - 5t\).

Substitute \(s = -3 - 5t\) into equation 3:

\(-4 + 3(-3 - 5t) = -2 - 4t\)

\(-4 - 9 - 15t = -2 - 4t\)

\(-13 - 15t = -2 - 4t\)

\(-11 = 11t\)

\(t = -1\).

Substitute \(t = -1\) into \(s = -3 - 5t\):

\(s = -3 - 5(-1) = 2\).

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

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

\(7 = p - 2\)

\(p = 9\).

Thus, the value of \(p\) is 9. The intersection point is:

\(\begin{pmatrix} 5 + 2 \\ 1 - 2 \\ -4 + 3(2) \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \\ 2 \end{pmatrix}\).

(a) To find the vector equation for the line passing through A and B, use the position vectors of A and B:

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

Equate the components of the general points on their line and \(l\):

\(\begin{align*} 1 + \lambda &= 1 + 2\mu, \\ 2 - 3\lambda &= -1 - 3\mu, \\ -2 + 3\lambda &= 3 + 4\mu. \end{align*}\)

Solving these equations gives \(\lambda = -1\) and \(\mu = -2\). However, substituting these values into the third equation does not satisfy it, confirming the lines do not intersect.

(b) For a general point \(P\) on \(l\), \(\mathbf{P} = -3\mathbf{j} + 5\mathbf{k} + \mu(2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k})\).

Calculate the scalar product of \(\overrightarrow{AP}\) and a direction vector for \(l\) and equate to zero:

\(4\mu + (9 + 9\mu) + (20 + 16\mu) = 0\)

Solving gives \(\mu = -1\).

Substitute \(\mu = -1\) back into the equation for \(P\) to find the position vector:

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

(i) Express the general point of l or m in component form: l: (3 - λ, -2 + 2λ, 1 + λ) and m: (4 + aμ, 4 + bμ, 2 - μ).

\(Equate components and eliminate either λ or μ from a pair of equations. For example, equate the i components: 3 - λ = 4 + aμ.\)

Eliminate the other parameter and obtain an equation in a and b. For example, equate the k components: 1 + λ = 2 - μ.

\(From these, solve to obtain 2a - b = 4.\)

\((ii) Using the correct process, equate the scalar product of the direction vectors to zero: (-1, 2, 1) ⋅ (a, b, -1) = 0.\)

\(This gives -a + 2b - 1 = 0.\)

\(Solve the simultaneous equations for a and b: 2a - b = 4 and -a + 2b - 1 = 0.\)

\(Obtain a = 3, b = 2.\)

(iii) Substitute found values in component equations and solve for λ or μ.

\(Obtain the position vector i + 2j + 3k from either λ = 2 or μ = -1.\)