(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:

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

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

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