Let the sum of the coefficient of first three terms in the expansion of $$\left(x - \frac{3}{x^2}\right)^n$$; $$x \neq 0$$, $$n \in \mathbb{N}$$ be 376. Then, the coefficient of $$x^4$$ is equal to:

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

The expansion is $$(x - 3x^{-2})^n$$. The general term is:

$$T_{r+1} = \binom{n}{r} (x)^{n-r} (-3x^{-2})^r = \binom{n}{r} (-3)^r x^{n-r-2r} = \binom{n}{r} (-3)^r x^{n-3r}$$

$$T_1 (r=0) = \binom{n}{0} (-3)^0 = 1$$

$$T_2 (r=1) = \binom{n}{1} (-3)^1 = -3n$$

$$T_3 (r=2) = \binom{n}{2} (-3)^2 = 9 \frac{n(n-1)}{2}$$

$$1 - 3n + \frac{9n(n-1)}{2} = 376$$

$$9n^2 - 15n - 750 = 0$$

Since $$n \in N$$, we take the positive root: $$n = 10$$.

$$10 - 3r = 4 \implies 3r = 6 \implies r = 2$$ (we need the term where the power of $$x$$ is 4 $$n - 3r = 4$$)

The coefficient is $$\binom{n}{r} (-3)^r$$: $$\text{Coeff} = \binom{10}{2} (-3)^2$$

$$\text{Coeff} = \frac{10 \times 9}{2} \times 9 = 45 \times 9 = 405$$