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

So both parts must be defined.

2. Domain from the inverse cosine term: $\cos^{-1}(u)$ is defined when $-1\le u\le 1.$

Here, $u=\frac{2-|x|}{4}.$ So, $-1\le \frac{2-|x|}{4}\le 1.$ Multiply throughout by $4$: $-4\le 2-|x|\le 4.$

Now solve both sides:

• From $-4\le 2-|x|,$ we get $-6\le -|x| \implies |x|\le 6.$

• From $2-|x|\le 4,$ we get $-|x|\le 2 \implies |x|\ge -2,$ which is always true.

Hence the condition from the first term is $|x|\le 6 \implies x\in[-6,6].$

3. Domain from the second term: $\left\{\ln(3-x)\right\}^{-1}=\frac{1}{\ln(3-x)}.$

For this to be defined:

• logarithm must exist: $3-x>0 \implies x<3,$
• denominator must be nonzero: $\ln(3-x)\ne 0.$

Now, $\ln(3-x)=0 \implies 3-x=1 \implies x=2.$ So we must exclude $x=2$.

Thus the condition from the second term is $x<3,\quad x\ne 2.$

4. Intersect both conditions:

From step 2: $x\in[-6,6]$

From step 3: $x<3$ and $x\ne 2$

Therefore, $\text{Domain}=[-6,3)-\{2\}.$

Comparing with the given form $[-\alpha,\beta)-\{\gamma\},$ we get $\alpha=6,\quad \beta=3,\quad \gamma=2.$

Hence, $\alpha+\beta+\gamma=6+3+2=11.$

5. Option check:

• A: $11$ ✅
• B: $12$
• C: $9$
• D: $8$

So the correct option is A.