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

Given:

$y={\cos }^{-1}\left|\frac{2-\left|x\right|}{4}\right|+\log {\left(3-x\right)}^{-1}$

$\Rightarrow -1\le \left|\frac{2-\left|x\right|}{4}\right|\le 1$

$\Rightarrow -4\le 2-\left|x\right|\le 4$

$\Rightarrow -6\le -\left|x\right|\le 2$

$\Rightarrow -2\le \left|x\right|\le 6$

$\Rightarrow \left|x\right|\le 6$

$\Rightarrow -6\le x\le 6 ...\left(i\right)$

Also, $3-x>0 \text{and} 3-x\ne 1$

$\Rightarrow x\ne 2 ...\left(ii\right)$ and $x<3 ...\left(iii\right)$

Using $\left(i\right), \left(ii\right) \text{and} \left(iii\right)$ we get,

$\Rightarrow x\in [-6,3)-\left\{2\right\}$

So, on comparing with $x\in [-\alpha ,\beta )-\left\{\gamma \right\}$ we get,

$\Rightarrow \alpha +\beta +\gamma =6+3+2=11$