(i) The sum of the first \(n\) terms of an arithmetic progression is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).

For the first 9 terms, \(S_9 = \frac{9}{2} (24 + 8d) = 135\).

Solving for \(d\), we have:

\(9(24 + 8d) = 270\)

\(24 + 8d = 30\)

\(8d = 6\)

\(d = \frac{3}{4}\).

(ii) The 9th term of the arithmetic progression is \(12 + 8 \times \frac{3}{4} = 18\).

In the geometric progression, the first term is 12, the second term is 18.

The common ratio \(r\) is \(\frac{18}{12} = \frac{3}{2}\).

The third term of the geometric progression is \(ar^2 = 12 \times \left(\frac{3}{2}\right)^2 = 27\).

The \(n\)th term of the arithmetic progression is \(12 + (n-1) \times \frac{3}{4} = 27\).

Solving for \(n\), we have:

\(12 + \frac{3}{4}(n-1) = 27\)

\(\frac{3}{4}(n-1) = 15\)

\(n-1 = 20\)

\(n = 21\).