(i) To find the midpoint of AB, use the midpoint formula:

\(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)

Substitute \(A(2, 5)\) and \(B(5, -1)\):

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

(ii) First, find the slope of AB:

\(m = \frac{-1 - 5}{5 - 2} = \frac{-6}{3} = -2\)

The slope of the line perpendicular to AB is the negative reciprocal:

\(m_{\perp} = \frac{1}{2}\)

Using point C(8, 6) and the point-slope form:

\(y - 6 = \frac{1}{2}(x - 8)\)

Simplify to get the equation in the form \(ax + by + c = 0\):

\(y - 6 = \frac{1}{2}x - 4\)

\(2y - 12 = x - 8\)

\(x - 2y + 4 = 0\)

First, solve the system of equations:

1. \(y^2 = 12x\)

2. \(3y = 4x + 6\)

From equation (2), express \(x\) in terms of \(y\):

\(x = \frac{3y - 6}{4}\)

Substitute \(x\) in equation (1):

\(y^2 = 12 \left( \frac{3y - 6}{4} \right)\)

Simplify:

\(y^2 = 9y - 18\)

Rearrange to form a quadratic equation:

\(y^2 - 9y + 18 = 0\)

Factor the quadratic:

\((y - 3)(y - 6) = 0\)

Thus, \(y = 3\) or \(y = 6\).

Find corresponding \(x\) values:

For \(y = 3\):

\(x = \frac{3(3) - 6}{4} = \frac{9 - 6}{4} = \frac{3}{4}\)

For \(y = 6\):

\(x = \frac{3(6) - 6}{4} = \frac{18 - 6}{4} = 3\)

The points of intersection are \(\left( \frac{3}{4}, 3 \right)\) and \((3, 6)\).

Calculate the distance between the points:

\(\text{Distance} = \sqrt{\left( 3 - \frac{3}{4} \right)^2 + (6 - 3)^2}\)

\(= \sqrt{\left( \frac{9}{4} \right)^2 + 3^2}\)

\(= \sqrt{\frac{81}{16} + 9}\)

\(= \sqrt{\frac{81}{16} + \frac{144}{16}}\)

\(= \sqrt{\frac{225}{16}}\)

\(= \frac{15}{4}\)

\(= 3.75\)

To solve the problem, we follow these steps:

1. Find the midpoint \(M\) of \(AB\):

\(M = \left( \frac{2 + 8}{2}, \frac{6 + 10}{2} \right) = (5, 8)\)

2. Calculate the gradient of \(AB\):

\(\text{Gradient of } AB = \frac{10 - 6}{8 - 2} = \frac{2}{3}\)

3. Find the gradient of the perpendicular bisector:

\(\text{Perpendicular gradient} = -\frac{3}{2}\)

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

\(y - 8 = -\frac{3}{2}(x - 5)\)

Expanding and rearranging gives:

\(2y + 3x = 31\)

5. Find the equation of line \(BC\):

\(y = 5(x - 6) \Rightarrow y = 5x - 30\)

6. Solve the system of equations:

\(2y + 3x = 31\)

\(y = 5x - 30\)

Substitute \(y = 5x - 30\) into \(2y + 3x = 31\):

\(2(5x - 30) + 3x = 31\)

\(10x - 60 + 3x = 31\)

\(13x = 91\)

\(x = 7\)

Substitute \(x = 7\) back into \(y = 5x - 30\):

\(y = 5(7) - 30 = 5\)

Thus, the coordinates of \(D\) are \((7, 5)\).

(i) To find the intersection points \(A\) and \(B\), solve the equations simultaneously:

\(x^2 - 4x + 7 = 9 - 3x\)

\(x^2 - x - 2 = 0\)

Solving this quadratic equation gives \(x = 2\) or \(x = -1\).

For \(x = 2\), \(y = 3\). For \(x = -1\), \(y = 12\).

Thus, \(A = (2, 3)\) and \(B = (-1, 12)\).

The midpoint \(M\) is:

\(M = \left( \frac{2 + (-1)}{2}, \frac{3 + 12}{2} \right) = \left( \frac{1}{2}, \frac{15}{2} \right)\)

(ii) The derivative of the curve is:

\(\frac{dy}{dx} = 2x - 4\)

Set \(\frac{dy}{dx} = -3\) (slope of the line):

\(2x - 4 = -3\)

\(2x = 1\)

\(x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into the curve equation:

\(y = \left( \frac{1}{2} \right)^2 - 4 \left( \frac{1}{2} \right) + 7 = \frac{1}{4} - 2 + 7 = \frac{5}{4}\)

So, \(Q = \left( \frac{1}{2}, \frac{5}{4} \right)\).

(iii) The distance \(MQ\) is:

\(MQ = \sqrt{\left( \frac{1}{2} - \frac{1}{2} \right)^2 + \left( \frac{7}{2} - \frac{5}{4} \right)^2}\)

\(MQ = \sqrt{0 + \left( \frac{14}{4} - \frac{5}{4} \right)^2}\)

\(MQ = \sqrt{\left( \frac{9}{4} \right)^2} = \frac{9}{4}\)

(i) Finding the coordinates of the intersection points:

Substitute \(y = 9 - \frac{6}{x}\) into \(y + x = 8\):

\(9 - \frac{6}{x} + x = 8\)

Simplify to find:

\(x^2 + x - 6 = 0\)

Solving the quadratic equation:

\((x - 2)(x + 3) = 0\)

Thus, \(x = 2\) or \(x = -3\).

For \(x = 2\), \(y = 9 - \frac{6}{2} = 6\).

For \(x = -3\), \(y = 9 - \frac{6}{-3} = 11\).

So, the points are \((2, 6)\) and \((-3, 11)\).

(ii) Finding the equation of the perpendicular bisector:

Midpoint of \((2, 6)\) and \((-3, 11)\):

\(\left( \frac{2 + (-3)}{2}, \frac{6 + 11}{2} \right) = \left( -\frac{1}{2}, \frac{17}{2} \right)\)

Gradient of the line joining the points:

\(m = \frac{11 - 6}{-3 - 2} = -1\)

Gradient of the perpendicular bisector:

\(m = 1\)

Equation of the perpendicular bisector:

\(y - \frac{17}{2} = 1 \left( x + \frac{1}{2} \right)\)

Simplify to:

\(y = x + 9\)