(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 4 - x\):

\(y = \int (4 - x) \, dx = 4x - \frac{1}{2}x^2 + c\).

Using the point \(P(2, 9)\), substitute to find \(c\):

\(9 = 4(2) - \frac{1}{2}(2)^2 + c\)

\(9 = 8 - 2 + c\)

\(c = 3\).

Thus, the equation of the curve is \(y = 4x - \frac{1}{2}x^2 + 3\).

(ii) The gradient of the tangent at \(P\) is \(\frac{dy}{dx} = 4 - 2 = 2\).

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

The equation of the normal is \(y - 9 = -\frac{1}{2}(x - 2)\).

(iii) Substitute \(y = 4x - \frac{1}{2}x^2 + 3\) into the normal equation:

\(4x - \frac{1}{2}x^2 + 3 - 9 = -\frac{1}{2}(x - 2)\)

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

\(x^2 - 9x + 14 = 0\)

Solving the quadratic equation gives \(x = 7\).

Substitute \(x = 7\) back into the curve equation:

\(y = 4(7) - \frac{1}{2}(7)^2 + 3\)

\(y = 28 - 24.5 + 3\)

\(y = 6.5\).

Thus, the coordinates of \(Q\) are \((7, 6.5)\).