Expansions!

Assuming no combination/cancellation, the expansion must have (9 + 1) + (12 + 1) = 23 terms. So two terms are missing. Either they combine with each other, or just cancel out.

Each term in $(2a^2 + 3b)^9$ has the form $\binom{9}{k_1}(2a^2)^{9-k_1}(3b)^{k_1}=\binom{9}{k_1}2^{9-k_1}3^{k_1}\cdot a^{18-2k_1}b^{k_1}$ while for $-k(3a+2b)^{12}$ , $-k\binom{12}{k_2}(3a)^{12-k_2}(2b)^{k_2}=-k\binom{12}{k_2}3^{12-k_2}2^{k_2}\cdot a^{12-k_2}b^{k_2}$

If terms are to combine or cancel out, their literal coefficients must be the same. Comparison gives $18-2k_1=12-k_2$ and $k_1=k_2$ , giving $k_1=k_2=6$ This is the only solution. Therefore, the two terms must have literal coefficient $a^6 b^6$ , and must cancel out. These terms are $\binom{9}{6}(2a^2)^{9-6}(3b)^{6}=489888a^6 b^6$ and $-k\binom{12}{6}(3a)^{12-6}(2b)^{6}=-43110144ka^6b^6$ Therefore, $489888a^6 b^6-43110144ka^6b^6=0\Longrightarrow k=\frac{1}{88}$ So $a+b=1+88=\boxed{89}$ .