To find the perpendicular distance from point P to the line l, we can use the vector approach.

First, find the vector PA where A is a point on the line. Let A be the point corresponding to \(\lambda = 0\), so \(A = \begin{pmatrix} 1 \\ 3 \\ -4 \end{pmatrix}\).

Then, \(\mathbf{PA} = \begin{pmatrix} 1 - (-1) \\ 3 - 4 \\ -4 - 11 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -15 \end{pmatrix}\).

The direction vector of the line is \(\mathbf{d} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\).

Use the cross product to find the area of the parallelogram formed by PA and d:

\(\mathbf{PA} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & -15 \\ 2 & 1 & 3 \end{vmatrix} = \begin{pmatrix} (-1)(3) - (-15)(1) \\ -((2)(3) - (-15)(2)) \\ (2)(1) - (-1)(2) \end{pmatrix} = \begin{pmatrix} 12 \\ -36 \\ 4 \end{pmatrix}\).

The magnitude of this cross product is \(\sqrt{12^2 + (-36)^2 + 4^2} = \sqrt{144 + 1296 + 16} = \sqrt{1456}\).

The magnitude of the direction vector \(\mathbf{d}\) is \(\sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}\).

The perpendicular distance is given by \(\frac{\sqrt{1456}}{\sqrt{14}} = \sqrt{104} \approx 10.2\).

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

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

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

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

Thus, \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) + (2\lambda \mathbf{i} + 2\lambda \mathbf{j} - 2\lambda \mathbf{k})\).

Therefore, \(\overrightarrow{OP} = (1 + 2\lambda)\mathbf{i} + (2 + 2\lambda)\mathbf{j} + (2 - 2\lambda)\mathbf{k}\).

(ii) To find \(\lambda\) such that \(OP\) bisects \(\angle AOB\), equate the expressions for \(\cos AOP\) and \(\cos BOP\):

\(\frac{9 + 2\lambda}{3\sqrt{9 + 4\lambda + 12\lambda^2}} = \frac{11 + 14\lambda}{5\sqrt{9 + 4\lambda + 12\lambda^2}}\)

Solving this equation gives \(\lambda = \frac{3}{8}\).

(iii) Verify \(AP : PB = OA : OB\):

\(AP = \lambda \times AB = \frac{3}{8} \times \sqrt{12}\)

\(PB = (1 - \lambda) \times AB = \frac{5}{8} \times \sqrt{12}\)

\(OA = \sqrt{9}\), \(OB = \sqrt{25}\)

\(\frac{AP}{PB} = \frac{3}{5} = \frac{OA}{OB}\)

Thus, the verification holds.

(i) To prove that the lines l and m do not intersect, express the general point of l in component form: \((2 + \lambda, -\lambda, 1 + 2\lambda)\) and the general point of m as \((\mu, 2 + 2\mu, 6 - 2\mu)\). Equate the components and solve for \(\lambda\) and \(\mu\). Possible solutions for \(\lambda\) are -2, \(\frac{1}{4}\), 7, and for \(\mu\) are 0, \(\frac{1}{4}\), -\(\frac{1}{2}\). Verify that all three component equations are not satisfied simultaneously, confirming that the lines do not intersect.

Alternatively, use the scalar triple product: \((2i - 2j - 5k) \cdot ((i - j + 2k) \times (i + 2j - 2k))\). Calculate the determinant to obtain a non-zero value, e.g., -27, indicating the lines do not intersect.

(ii) To find the acute angle between the directions of l and m, calculate the scalar product of the direction vectors \((i - j + 2k)\) and \((i + 2j - 2k)\). The scalar product is -3. Calculate the moduli of the direction vectors: \(\sqrt{6}\) and \(\sqrt{9}\). Use the formula \(\cos \theta = \frac{-3}{\sqrt{6} \times \sqrt{9}}\) to find \(\theta\). The acute angle is approximately 47.1° or 0.822 radians.

(i) The direction vector of line AB is \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{i} + 4\mathbf{j}) - (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\).

Thus, the vector equation of the line AB is \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda(2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})\).

(ii) Since OP is perpendicular to AB, the dot product \(\overrightarrow{OP} \cdot \overrightarrow{AB} = 0\).

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

Then, \(\overrightarrow{OP} \cdot \overrightarrow{AB} = (1 + 2\lambda)(2) + (2 + 2\lambda)(2) + (2 - 2\lambda)(-2) = 0\).

Simplifying, \(2 + 4\lambda + 4 + 4\lambda - 4 + 4\lambda = 0\).

Thus, \(12\lambda + 2 = 0\), giving \(\lambda = -\frac{1}{6}\).

Substitute \(\lambda = -\frac{1}{6}\) into \(\overrightarrow{OP}\):

\(\overrightarrow{OP} = (1 + 2(-\frac{1}{6}))\mathbf{i} + (2 + 2(-\frac{1}{6}))\mathbf{j} + (2 - 2(-\frac{1}{6}))\mathbf{k}\).

\(\overrightarrow{OP} = \frac{2}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} + \frac{7}{3}\mathbf{k}\).

(i) Express the general point of l in component form: \((1 + s, 1 - s, 1 + 2s)\).

Express the general point of m in component form: \((4 + 2t, 6 + 2t, 1 + t)\).

Equate the components:

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

\(1 - s = 6 + 2t\)

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

Solving these equations, we find \(s = -1\) or \(t = -2\).

Verify that all three component equations are satisfied with these values.

(ii) The direction vector of l is \((1, -1, 2)\) and of m is \((2, 2, 1)\).

Calculate the scalar product: \(1 \times 2 + (-1) \times 2 + 2 \times 1 = 2 - 2 + 2 = 2\).

Calculate the moduli: \(\sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6}\) and \(\sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3\).

Divide the scalar product by the product of the moduli: \(\frac{2}{\sqrt{6} \times 3}\).

Evaluate the inverse cosine: \(\cos^{-1}\left(\frac{2}{3\sqrt{6}}\right)\).

The acute angle between the lines is \(74.2^\circ\) (or 1.30 radians).