(i) The perimeter of the sector is given by \(r + r + r\theta = 24\), which simplifies to \(2r + r\theta = 24\). Solving for \(\theta\), we get \(\theta = \frac{24 - 2r}{r}\).

The area of the sector is \(A = \frac{1}{2} r^2 \theta\). Substituting \(\theta\), we have:

\(A = \frac{1}{2} r^2 \left( \frac{24 - 2r}{r} \right) = \frac{24r}{2} - \frac{r^2}{2} = 12r - r^2\).

(ii) To express \(A\) in the form \(a - (r - b)^2\), we complete the square:

\(A = 12r - r^2 = -(r^2 - 12r)\).

Completing the square gives \(-(r^2 - 12r) = -( (r - 6)^2 - 36 ) = 36 - (r - 6)^2\).

Thus, \(A = 36 - (r - 6)^2\).

(iii) The expression \(A = 36 - (r - 6)^2\) is maximized when \((r - 6)^2 = 0\), which occurs when \(r = 6\).

Substituting \(r = 6\) into the perimeter equation \(2r + r\theta = 24\), we get:

\(12 + 6\theta = 24\) which simplifies to \(\theta = 2\).

Therefore, the greatest value of \(A\) is 36 when \(r = 6\) and \(\theta = 2\).