Let the sum of the coefficients of the first three terms in the expansion of ${\left(\mathrm{x}-\frac{3}{{\mathrm{x}}^2}\right)}^{\mathrm{n}},\mathrm{x}\ne 0,\mathrm{n}\in \mathrm{N}$ , be 376. Then the coefficient of  $x^4$ is 

Given binomial expansion is 
Sum of the coefficient of first 3 terms ${\left(\mathrm{x}-\frac{3}{{\mathrm{x}}^2}\right)}^{\mathrm{n}}$
$\begin{array}{l}{ }^{\mathrm{n}}{\mathrm{C}}_0{-}^{\mathrm{n}}{\mathrm{C}}_1\left(3\right){+}^{\mathrm{n}}{\mathrm{C}}_2\cdot 3^2=372\\ \Rightarrow 1-3\mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-1)}{1-2}\cdot 9=372\\ \Rightarrow 9{\mathrm{n}}^2-15\mathrm{n}-750=0\\ \Rightarrow (\mathrm{n}-10)(9\mathrm{n}+75)=0\\ \therefore \mathrm{n}=10\end{array}$
Now  ${\mathrm{T}}_{2+1}{=}^{10}{\mathrm{C}}_2{\mathrm{x}}^8{\left(-\frac{3}{{\mathrm{x}}^2}\right)}^2{=}^{10}{\mathrm{C}}_2\times 9\times {\mathrm{x}}^4$
 Coefficient $x^4\text{ in }\left(x-\frac{3}{x^2}\right){=}^{10}{C}_2\times 9=405$