The coefficient of $$x^5$$ in the expansion of $$\left(2x^3 - \frac{1}{3x^2}\right)^5$$ is

The general term in the expansion of $$\left(2x^3 - \frac{1}{3x^2}\right)^5$$ is:

$$T_{r+1} = \binom{5}{r}(2x^3)^{5-r}\left(-\frac{1}{3x^2}\right)^r = \binom{5}{r}2^{5-r}\left(-\frac{1}{3}\right)^r x^{3(5-r) - 2r}$$

$$= \binom{5}{r}2^{5-r}\left(-\frac{1}{3}\right)^r x^{15-5r}$$

For coefficient of $$x^5$$: $$15 - 5r = 5$$, so $$r = 2$$.

$$T_3 = \binom{5}{2} \cdot 2^3 \cdot \left(-\frac{1}{3}\right)^2 \cdot x^5$$

$$= 10 \times 8 \times \frac{1}{9} \times x^5 = \frac{80}{9}x^5$$

The coefficient of $$x^5$$ is $$\frac{80}{9}$$.

This matches option 1.