If the constant term in the expansion of $$\left(1 + 2x - 3x^3\right)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$$ is $$p$$, then $$108p$$ is equal to ______

Find the constant term in $$(1 + 2x - 3x^3)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$$.

General term of $$\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$$:

$$ T_{r+1} = \binom{9}{r}\left(\frac{3}{2}x^2\right)^{9-r}\left(-\frac{1}{3x}\right)^r = \binom{9}{r}\frac{3^{9-r}}{2^{9-r}} \cdot \frac{(-1)^r}{3^r} \cdot x^{2(9-r)-r} $$

$$ = \binom{9}{r}\frac{(-1)^r \cdot 3^{9-2r}}{2^{9-r}} x^{18-3r} $$

The constant term in the full product comes from:

1. $$1 \times$$ (term with $$x^0$$ from the expansion): $$18 - 3r = 0 \Rightarrow r = 6$$.

2. $$2x \times$$ (term with $$x^{-1}$$): $$18 - 3r = -1 \Rightarrow 3r = 19$$. Not integer. No contribution.

3. $$-3x^3 \times$$ (term with $$x^{-3}$$): $$18 - 3r = -3 \Rightarrow r = 7$$.

Contribution 1 ($$r = 6$$):

$$ 1 \times \binom{9}{6}\frac{(-1)^6 \cdot 3^{-3}}{2^3} = 84 \times \frac{1}{27 \times 8} = 84 \times \frac{1}{216} = \frac{84}{216} = \frac{7}{18} $$

Contribution 3 ($$r = 7$$):

$$ -3 \times \binom{9}{7}\frac{(-1)^7 \cdot 3^{-5}}{2^2} = -3 \times 36 \times \frac{-1}{243 \times 4} = -3 \times 36 \times \frac{-1}{972} = \frac{108}{972} = \frac{1}{9} $$

Total constant term:

$$ p = \frac{7}{18} + \frac{1}{9} = \frac{7}{18} + \frac{2}{18} = \frac{9}{18} = \frac{1}{2} $$

$$ 108p = 108 \times \frac{1}{2} = 54 $$

The answer is 54.