The coefficient of \(x^3\) in the expansion of \((3 + 2ax)^5\) is given by:

\(\binom{5}{3} (3)^{2} (2ax)^{3} = 10 \times 9 \times 8a^3 = 720a^3\)

The coefficient of \(x^2\) in the expansion of \((2 + ax)^6\) is given by:

\(\binom{6}{2} (2)^{4} (ax)^{2} = 15 \times 16 \times a^2 = 240a^2\)

According to the problem, the coefficient of \(x^3\) is six times the coefficient of \(x^2\):

\(720a^3 = 6 \times 240a^2\)

Simplifying gives:

\(720a^3 = 1440a^2\)

Dividing both sides by \(240a^2\) gives:

\(3a = 6\)

Thus, \(a = 2\).