Let $\left[x\right]$ denote the greatest integer $\le x$, where $x\in R$. If the domain of the real valued function $f\left(x\right)=\sqrt{\frac{\left|\left[x\right]\right|-2}{\left|\left[x\right]\right|-3}}$ is $\left(-\infty ,a\right)\cup \left[b,c\right)\cup \left[4,\infty \right),a<b<c$, then the value of $a+b+c$ is:

$f\left(x\right)=\sqrt{\frac{\left|\left[x\right]\right|-2}{\left|\left[x\right]\right|-3}}$

For domain,

$\left(\frac{\left|\left[x\right]\right|-2}{\left|\left[x\right]\right|-3}\right)\ge 0$

Case $\mathrm{I}:$ When $\left|\right[x\left]\right|-2\ge 0$ and $\left|\left[x\right]\right|-3>0$, then

$\therefore \mathrm{x}\in \left(-\infty ,-3\right)\cup \left[4,\infty \right) ...\left(1\right)$

Case $\mathrm{II}:$ When $\left|\left[x\right]\right|-2\le 0$ and $\left|\left[x\right]\right|-3<0$, then

$\therefore x\in \left[-2,3\right) ...\left(2\right)$

So, from $\left(1\right)$ and $\left(2\right)$, we get

Domain of function is

$\left(-\infty ,-3\right)\cup \left[-2,3\right)\cup \left[4,\infty \right)$

$\therefore \left(a+b+c\right)=-3+\left(-2\right)+3=-2$

$\left(\because a<b<c\right)$