If the domain of the function $$\log_{5}(18x - x^{2} - 77)$$ is $$(\alpha,\beta)$$ and the domain of the function $$\log_{(x-1)}\left(\frac{2x^{2}+3x-2}{x^{2}-3x-4}\right)$$ is $$(\gamma,\delta)$$, then $$\alpha^{2}+\beta^{2}+\gamma^{2}$$ is equal to :

Find $$\alpha^2 + \beta^2 + \gamma^2$$ where $$(\alpha, \beta)$$ is domain of $$\log_5(18x - x^2 - 77)$$ and $$(\gamma, \delta)$$ is domain of $$\log_{(x-1)}\left(\frac{2x^2+3x-2}{x^2-3x-4}\right)$$.

Domain of $$\log_5(18x - x^2 - 77)$$.

Need $$18x - x^2 - 77 > 0$$, i.e., $$x^2 - 18x + 77 < 0$$.

$$x^2 - 18x + 77 = (x-7)(x-11) < 0$$

Domain: $$(7, 11)$$, so $$\alpha = 7, \beta = 11$$.

Domain of $$\log_{(x-1)}\left(\frac{2x^2+3x-2}{x^2-3x-4}\right)$$.

Need: (i) $$x - 1 > 0$$ and $$x - 1 \neq 1$$: $$x > 1, x \neq 2$$.

(ii) $$\frac{2x^2+3x-2}{x^2-3x-4} > 0$$.

Factor: $$2x^2+3x-2 = (2x-1)(x+2)$$, $$x^2-3x-4 = (x-4)(x+1)$$.

$$\frac{(2x-1)(x+2)}{(x-4)(x+1)} > 0$$

Critical points: $$x = -2, -1, 1/2, 4$$. Combined with $$x > 1, x \neq 2$$:

For $$x > 1$$: check intervals $$(1, 4)$$ and $$(4, \infty)$$.

At $$x = 2$$: $$\frac{(3)(4)}{(-2)(3)} = \frac{12}{-6} = -2 < 0$$. Not in domain.

At $$x = 3$$: $$\frac{(5)(5)}{(-1)(4)} = \frac{25}{-4} < 0$$. Not in domain.

At $$x = 5$$: $$\frac{(9)(7)}{(1)(6)} > 0$$. In domain.

So domain for $$x > 1$$: $$(4, \infty) \setminus \{2\}$$. But 2 is not in $$(4, \infty)$$, so domain is $$(4, \infty)$$.

Wait, we also need to exclude $$x = 2$$ (base = 1). Since $$4 > 2$$, this is automatically excluded.

So $$\gamma = 4$$. But we need $$(\gamma, \delta)$$. If domain is $$(4, \infty)$$, then $$\delta = \infty$$. This doesn't work for the problem.

The answer is Option C: 186. So $$\alpha^2 + \beta^2 + \gamma^2 = 49 + 121 + 16 = 186$$. With $$\gamma = 4$$.

The correct answer is Option C: 186.