(i) To find the first three terms of \((2 + 3x)^6\), we use the binomial expansion formula:

\((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

For \((2 + 3x)^6\), \(a = 2\), \(b = 3x\), and \(n = 6\).

The first three terms are:

1st term: \(\binom{6}{0} 2^6 (3x)^0 = 64\)

2nd term: \(\binom{6}{1} 2^5 (3x)^1 = 192 \times 3x = 576x\)

3rd term: \(\binom{6}{2} 2^4 (3x)^2 = 15 \times 16 \times 9x^2 = 2160x^2\)

Thus, the first three terms are \(64 + 576x + 2160x^2\).

(ii) In the expansion of \((1 + ax)(2 + 3x)^6\), the coefficient of \(x^2\) is zero.

The term \(x^2\) comes from two sources:

1. The \(x^2\) term from \((2 + 3x)^6\), which is \(2160x^2\).

2. The product of the \(x\) term from \((1 + ax)\) and the \(x\) term from \((2 + 3x)^6\), which is \(576ax^2\).

Setting the sum of these coefficients to zero:

\(576a + 2160 = 0\)

Solving for \(a\):

\(576a = -2160\)

\(a = -\frac{2160}{576} = -\frac{15}{4}\)

Thus, \(a = -\frac{15}{4}\).