Let the domain of the function f(x) = $$\log_{3}\log_{5}(7-\log_{2}(x^{2}-10x+85))+\sin^{-1}\left(|\frac{3x-7}{17-x}|\right)$$ be $$(\alpha, \beta)$$. Then $$\alpha + \beta$$ is equal to :

The term inside the base-2 logarithm is$$x^{2}-10x+85=(x-5)^{2}+60$$ which is always positive for all $$x\in\mathbb{R}$$.

Hence $$\log_{2}(x^{2}-10x+85)$$ is defined for every real $$x$$.

Argument of the base-5 logarithm
For $$\log_{5}\bigl(7-\log_{2}(x^{2}-10x+85)\bigr)$$ we require

$$7-\log_{2}(x^{2}-10x+85) \gt 1$$

The inequality is strict because the result of the base-5 logarithm, $$\log_{5}(y)$$, must be positive (otherwise it would become the argument of another logarithm in Step 3).
Rewrite the inequality:

$$\log_{2}(x^{2}-10x+85) \lt 6$$

Converting to exponential form,

$$(x-5)^{2}+60 \lt 2^{6}=64$$

$$\Longrightarrow\;(x-5)^{2} \lt 4$$

$$\Longrightarrow\;-2 \lt x-5 \lt 2$$

$$\Longrightarrow\;3 \lt x \lt 7$$

Argument of the base-3 logarithm
Because $$\log_{5}(7-\log_{2}(\ldots))\gt 0$$ is already ensured in Step 2, the outer logarithm $$\log_{3}(\ldots)$$ is automatically defined on $$3\lt x\lt 7$$. No further restriction arises here.

Argument of $$\sin^{-1}$$
For $$\sin^{-1}\!\left(\left|\dfrac{3x-7}{17-x}\right|\right)$$ we need

$$\left|\dfrac{3x-7}{17-x}\right|\le 1$$ and also $$17-x\neq 0\;(\text{i.e.\ }x\neq 17).$$

The absolute-value inequality is equivalent to

$$(3x-7)^{2}\le(17-x)^{2}$$

Expanding both squares gives

$$9x^{2}-42x+49\;\le\;x^{2}-34x+289$$

$$\Longrightarrow\;8x^{2}-8x-240\le 0$$

Dividing by $$8$$ and factoring,

$$(x-6)(x+5)\le 0$$

$$\Longrightarrow\;-5\le x\le 6$$

Since $$x\neq 17$$ is already outside this range, it causes no extra restriction.

Intersection of all conditions

$$3\lt x\lt 7$$
$$-5\le x\le 6$$

The intersection is

$$3\lt x\le 6$$

Thus the domain can be written as $$(\alpha,\beta)=(3,6)$$. (The point $$x=6$$ satisfies every condition but the interval symbol chosen in the question is open; this does not affect the values of $$\alpha$$ and $$\beta$$.)

$$\alpha+\beta = 3+6 = 9$$