(i) To find the distance \(AB\), use the distance formula:

\(\sqrt{(9 - p)^2 + (3p)^2} = 13\)

Squaring both sides, we get:

\((9 - p)^2 + (3p)^2 = 169\)

Expanding and simplifying:

\(81 - 18p + p^2 + 9p^2 = 169\)

\(10p^2 - 18p + 81 = 169\)

\(10p^2 - 18p - 88 = 0\)

Solving this quadratic equation using the quadratic formula:

\(p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 10\), \(b = -18\), \(c = -88\):

\(p = \frac{18 \pm \sqrt{(-18)^2 - 4 \times 10 \times (-88)}}{20}\)

\(p = \frac{18 \pm \sqrt{324 + 3520}}{20}\)

\(p = \frac{18 \pm \sqrt{3844}}{20}\)

\(p = \frac{18 \pm 62}{20}\)

\(p = 4 \quad \text{or} \quad p = -\frac{11}{5}\)

(ii) The gradient of the line \(2x + 3y = 9\) is \(-\frac{2}{3}\).

The gradient of \(AB\) is \(\frac{3p}{9 - p}\).

For the lines to be perpendicular, the product of their gradients must be \(-1\):

\(\left(-\frac{2}{3}\right) \left(\frac{3p}{9 - p}\right) = -1\)

\(\frac{2p}{9 - p} = 1\)

Solving for \(p\):

\(2p = 9 - p\)

\(3p = 9\)

\(p = 3\)

(i) Find the equation of the perpendicular bisector of AB.

1. Calculate the midpoint M of AB:

\(M = \left( \frac{4 + 10}{2}, \frac{6 + 2}{2} \right) = (7, 4)\)

2. Find the slope of AB:

\(m_{AB} = \frac{2 - 6}{10 - 4} = -\frac{2}{3}\)

3. The slope of the perpendicular bisector is the negative reciprocal:

\(m_{\text{perpendicular}} = \frac{3}{2}\)

4. Use point-slope form to find the equation:

\(y - 4 = \frac{3}{2}(x - 7)\)

(ii) Calculate the coordinates of C.

1. Equation of line parallel to AB through (3, 11):

\(y - 11 = -\frac{2}{3}(x - 3)\)

2. Solve the system of equations:

Equation 1: \(y - 4 = \frac{3}{2}(x - 7)\)

Equation 2: \(y - 11 = -\frac{2}{3}(x - 3)\)

3. Substitute and solve:

\(\frac{3}{2}(x - 7) + 4 = -\frac{2}{3}(x - 3) + 11\)

4. Solve for x:

\(\frac{3}{2}x - \frac{21}{2} + 4 = -\frac{2}{3}x + 2 + 11\)

5. Simplify and solve:

\(\frac{3}{2}x + \frac{2}{3}x = \frac{21}{2} - 4 + 13\)

6. Find x:

\(x = 9\)

7. Substitute x back to find y:

\(y - 4 = \frac{3}{2}(9 - 7)\)

\(y = 7\)

Coordinates of C: (9, 7)

(i) The equation of the line with gradient \(-2\) passing through \(P(3t, 2t)\) is:

\(y - 2t = -2(x - 3t)\)

Rearranging gives:

\(y = -2x + 6t + 2t\)

\(y = -2x + 8t\)

To find \(A\), set \(y = 0\):

\(0 = -2x + 8t\)

\(x = 4t\)

So, \(A(4t, 0)\).

To find \(B\), set \(x = 0\):

\(y = 8t\)

So, \(B(0, 8t)\).

The area of triangle \(AOB\) is:

\(\text{Area} = \frac{1}{2} \times 4t \times 8t = 16t^2\)

(ii) The line through \(P\) perpendicular to \(AB\) has gradient \(\frac{1}{2}\). Its equation is:

\(y - 2t = \frac{1}{2}(x - 3t)\)

Rearranging gives:

\(y = \frac{1}{2}x - \frac{3}{2}t + 2t\)

\(y = \frac{1}{2}x + \frac{1}{2}t\)

To find \(C\), set \(y = 0\):

\(0 = \frac{1}{2}x + \frac{1}{2}t\)

\(x = -t\)

So, \(C(-t, 0)\).

The midpoint of \(PC\) is:

\(\left( \frac{3t - t}{2}, \frac{2t + 0}{2} \right) = (t, t)\)

This lies on the line \(y = x\).

(i) Find the gradient of line AB:

The gradient \(m\) of line \(AB\) is given by:

\(m = \frac{3a + 9 - (2a - 1)}{2a + 4 - a} = \frac{a + 10}{a + 4}\)

The gradient of a line perpendicular to \(AB\) is the negative reciprocal:

\(m_{\perp} = -\frac{a + 4}{a + 10}\)

(ii) Find the possible values of \(a\):

The distance \(AB\) is given by:

\(\sqrt{(2a + 4 - a)^2 + (3a + 9 - (2a - 1))^2} = \sqrt{260}\)

Simplify the expression:

\(\sqrt{(a + 4)^2 + (a + 10)^2} = \sqrt{260}\)

Square both sides:

\((a + 4)^2 + (a + 10)^2 = 260\)

Expand and simplify:

\(a^2 + 8a + 16 + a^2 + 20a + 100 = 260\) \(2a^2 + 28a + 116 = 260\) \(2a^2 + 28a - 144 = 0\)

Divide by 2:

\(a^2 + 14a - 72 = 0\)

Factorize:

\((a - 4)(a + 18) = 0\)

Thus, \(a = 4\) or \(a = -18\).

Step-by-step Solution:

1. Substitute point A (8, -4) into the line equation 4x + ky = 20:
   \(4(8) + k(-4) = 20\)
   \(32 - 4k = 20\)
   Solve for \(k\):
   \(32 - 20 = 4k\)
   \(12 = 4k\)
   \(k = 3\)
2. Substitute point B (b, 2b) into the line equation:
   \(4b + 3(2b) = 20\)
   \(4b + 6b = 20\)
   \(10b = 20\)
   Solve for \(b\):
   \(b = 2\)
3. Find the mid-point of AB:
   Mid-point formula: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
   \(A (8, -4), B (2, 4)\)
   Mid-point: \(\left( \frac{8 + 2}{2}, \frac{-4 + 4}{2} \right) = (5, 0)\)