If $$\mathop {\lim }\limits_{x \to 0} {{\alpha x{e^x} - \beta {{\log }_e}(1 + x) + \gamma {x^2}{e^{ - x}}} \over {x{{\sin }^2}x}} = 10,\alpha ,\beta ,\gamma \in R$$, then the value of $\alpha$ + $\beta$ + $\gamma$ is _____________.

$$\mathop {\lim }\limits_{x \to 0} {{\alpha x\left( {1 + x + {{{x^2}} \over x}} \right) - \beta \left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3}} \right) + \gamma {x^2}(1 - x)} \over {{x^3}}}$$<br><br>$$\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) + {x^3}\left( {{\alpha \over 2} - {\beta \over 3} - \gamma } \right)} \over {{x^3}}} = 10$$<br><br>For limit to exist<br><br>$\alpha - \beta = 0,\alpha + {\beta \over 2} + \gamma = 0$${\alpha \over 2} - {\beta \over 3} - \gamma = 10$ ..... (i)<br><br>$\beta = \alpha ,\gamma = - 3{\alpha \over 2}$<br><br>Put in (i)<br><br>${\alpha \over 2} - {\alpha \over 3} + {{3\alpha } \over 2} = 10$<br><br>$${\alpha \over 6} + {{3\alpha } \over 2} = 10 \Rightarrow {{\alpha + 9\alpha } \over 6} = 10$$<br><br>$\Rightarrow \alpha = 6$<br><br>$\alpha$ = 6, $\beta$ = 6, $\gamma$ = $-$9<br><br>$\alpha$ + $\beta$ + $\gamma$ = 3