1. Let $n=[x]$ where $n$ is an integer.

Then the function becomes $f(x)=\sqrt{\frac{|n|-2}{|n|-3}}.$ For the function to be defined, we need:

• the denominator nonzero: $|n|-3\ne 0 \implies |n|\ne 3$
• the quantity inside the square root non-negative: $\frac{|n|-2}{|n|-3}\ge 0.$

So put $t=|n|,$ where $t$ is a non-negative integer. Then we solve $\frac{t-2}{t-3}\ge 0, \quad t\ne 3.$

2. Solve the inequality: Critical points are $t=2$ and $t=3$.

Check intervals:

• For $0\le t<2$: numerator $<0$, denominator $<0$, so ratio $>0$.
• For $t=2$: ratio $=0$, allowed.
• For $2<t<3$: numerator $>0$, denominator $<0$, so ratio $<0$.
• For $t>3$: numerator $>0$, denominator $>0$, so ratio $>0$.

Hence allowed values are $t\in [0,2]\cup(3,\infty).$ Since $t=|n|$ is integer, this means $|n|\in\{0,1,2,4,5,6,\dots\}.$ So forbidden values are only $|n|=3.$ Thus $n\ne \pm 3.$

3. Since $n=[x]$, the domain excludes all $x$ such that $[x]=3 \quad \text{or} \quad [x]=-3.$ Now,

• $[x]=3$ for $x\in [3,4),$
• $[x]=-3$ for $x\in [-3,-2).$

Therefore the domain is $\mathbb R\setminus\big([-3,-2)\cup[3,4)\big).$ So, $\text{Domain}=(-\infty,-3)\cup[-2,3)\cup[4,\infty).$

4. Compare with the given form $(-\infty,a)\cup[b,c)\cup[4,\infty), \quad a<b<c.$ Thus, $a=-3,\quad b=-2,\quad c=3.$ Hence, $a+b+c=-3+(-2)+3=-2.$

5. Therefore the correct option is $\boxed{-2}$ which is option C.