If $\underset{x\to 0}{\lim }\frac{\alpha e^x+\beta e^{-x}+\gamma \sin x}{x{\sin }^{2}x}=\frac{2}{3}$ where $\alpha ,\beta ,\gamma \in R$, then which of the following is/are correct

$\underset{x\to 0}{\lim }\frac{\alpha \left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+--\right)+\beta \left(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+---\right)+\gamma \left(x-\frac{x^3}{3!}+\frac{x^5}{5!}--\right)}{x^3}$

$\alpha +\beta =0,\alpha -\beta +\gamma =0$

and $\underset{x\to 0}{\lim }\frac{x^3\left(\frac{\alpha }{3!}-\frac{\beta }{3!}-\frac{\gamma }{3!}\right)+x^4\left(-\frac{\alpha }{3!}-\frac{\beta }{3!}-\frac{\gamma }{3!}\right)}{x^3}=\frac{2}{3}$

$\begin{array}{l}\frac{\alpha }{6}-\frac{\beta }{6}-\frac{\gamma }{6}=\frac{2}{3}\Rightarrow \alpha -\mathit{\beta }\mathbf{-}\mathit{\gamma }\mathbf{=}\mathbf{4}\\ \Rightarrow -2\gamma =4\Rightarrow \gamma =-2\\ \therefore \alpha +\beta =0,\alpha -\beta =2\\ \Rightarrow \alpha =1,\beta =-1,\gamma =-2\\ {\alpha }^2+{\beta }^2+{\gamma }^2=6,{\alpha }^2-{\beta }^2+{\gamma }^2=4,\alpha \beta +\beta \gamma +\gamma \alpha =-1+2-2=-1\end{array}$