(i) Differentiate the curve: \(\frac{dy}{dx} = 4 - 2x\).

At \(x = 3\), the gradient \(m = -2\).

The gradient of the normal is \(\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

Equation of the normal: \(y - 6 = \frac{1}{2}(x - 3)\).

Simplify to get \(2y = x + 9\).

(ii) The normal meets the axes at \((0, \frac{9}{2})\) and \((-9, 0)\).

Mid-point formula: \(\left(\frac{-9 + 0}{2}, \frac{0 + \frac{9}{2}}{2}\right) = \left(-\frac{9}{2}, \frac{9}{4}\right)\).

(iii) Set \(2y = x + 9\) equal to \(y = 4x - x^2 + 3\).

Substitute to get \(2(4x - x^2 + 3) = x + 9\).

Simplify to \(2x^2 - 7x + 3 = 0\).

Solving gives \(x = \frac{1}{2}\).

Substitute back to find \(y = \frac{43}{4}\).

Coordinates are \(\left(\frac{1}{2}, \frac{43}{4}\right)\).