If the domain of the function $$ \large f(x)=\sin ^{-1} \left( \frac{5-x}{3+2x} \right)+\frac{1}{\log_{e}{(10-x)}} $$ is $$ \large (-\infty,\propto] \cup [\beta,\gamma) - \left\{ \delta\right\} $$, then $$ \large 6(\alpha+ \beta+ \gamma+\delta) $$ is equal to

We need the domain of
$$f(x)=\sin^{-1}\left(\frac{5-x}{3+2x}\right)+\frac{1}{\log_e(10-x)}$$

For ($$\sin^{-1})$$, the argument must satisfy
$$-1\le\frac{5-x}{3+2x}\le1$$

$$Also(3+2x\ne0\Rightarrow x\ne-\frac{3}{2}).$$

Solve:
$$\frac{5-x}{3+2x}\le1$$

$$\frac{5-x-(3+2x)}{3+2x}\le0$$
$$\frac{2-3x}{3+2x}\le0$$

Critical points:

$$(x=\frac{2}{3},-\frac{3}{2}).$$

This gives
$$x\in(-\infty,-\frac{3}{2})\cup[\frac{2}{3},\infty)$$

Now solve
$$\frac{5-x}{3+2x}\ge-1$$

$$\frac{5-x+3+2x}{3+2x}\ge0$$

$$\frac{x+8}{3+2x}\ge0$$

Critical points: (x=-8,-\frac32).

Hence
$$x\in(-\infty,-8]\cup(-\frac{3}{2},\infty)$$
$$x\in(-\infty,-8]\cup[\frac{2}{3},\infty)$$
$$\frac{1}{\log_e(10-x)}$$
we need

1. $$(10-x>0\Rightarrow x<10)$$
2. $$(\log_e(10-x)\ne0\Rightarrow10-x\ne1\Rightarrow x\ne9)$$

Therefore domain:
$$(-\infty,-8]\cup\left[\frac{2}{3},10\right)-9$$

Comparing with
$$(-\infty,\alpha]\cup[\beta,\gamma)-\delta$$

we get
$$\alpha=-8,\quad\beta=\frac{2}{3},\quad\gamma=10,\quad\delta=9$$
$$6(\alpha+\beta+\gamma+\delta)$$

$$=6\left(-8+\frac{2}{3}+10+9\right)$$
$$=6\left(11+\frac{2}{3}\right)$$
$$=6\cdot\frac{35}{3}$$
=70