If the domain of the function $$f(x) = \cos^{-1}\left(\frac{2 - |x|}{4}\right) + (\log_e(3 - x))^{-1}$$ is $$[-\alpha, \beta) - \{\gamma\}$$, then $$\alpha + \beta + \gamma$$ is equal to :

We need to find the domain of $$f(x) = \cos^{-1}\left(\frac{2-|x|}{4}\right) + (\log_e(3-x))^{-1}$$.

For $$\cos^{-1}\left(\frac{2-|x|}{4}\right)$$ to be defined we require $$-1 \leq \frac{2-|x|}{4} \leq 1$$. The left inequality gives $$2 - |x| \geq -4$$ so $$|x| \leq 6$$, i.e., $$-6 \leq x \leq 6$$, whereas the right inequality gives $$2 - |x| \leq 4$$ so $$|x| \geq -2$$ which is always true. Thus this condition yields $$x \in [-6, 6]$$.

For $$(\log_e(3-x))^{-1}$$ to be defined we need $$3 - x > 0$$ (so $$x < 3$$) and $$\log_e(3-x) \neq 0$$ which implies $$3 - x \neq 1$$, i.e., $$x \neq 2$$.

Intersecting these gives $$x \in [-6,6] \cap (-\infty,3)\setminus\{2\} = [-6,3)\setminus\{2\}$$, so the domain is $$[-6,3) - \{2\}$$, which means $$\alpha = 6$$, $$\beta = 3$$, $$\gamma = 2$$, and therefore $$\alpha + \beta + \gamma = 6 + 3 + 2 = 11$$.

The correct answer is Option 3: 11.