Three constraints:

1) ${\sin }^{-1}$ domain: $\left|\frac{5-x}{3+2x}\right|\le 1$, i.e. $(5-x{)}^2\le (2x+3{)}^2$. Factor as a difference of squares: $[(2x+3)-(5-x)][(2x+3)+(5-x)]\ge 0$, i.e. $(3x-2)(x+8)\ge 0$, giving $x\le -8$ or $x\ge 2/3$.

2) ${\log }_e(10-x)$ defined: $10-x>0$, so $x<10$.

3) Denominator ${\log }_e(10-x)\ne 0$: $10-x\ne 1$, so $x\ne 9$.

Combining: $x\in (-\infty ,-8]\cup [2/3,10)-\{9\}$.

So $\alpha =-8,\beta =2/3,\gamma =10,\delta =9$. Then $6(\alpha +\beta +\gamma +\delta )=6(-8+2/3+10+9)=6\cdot \frac{35}{3}=70$.