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

For domain,<br><br>${{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$ $\ge$ 0<br><br>Case I :<br><br>When ${\left| {[x]} \right| - 2}$ $\ge$ 0<br><br>and ${\left| {[x]} \right| - 3}$ &gt; 0<br><br>$\therefore$ x $\in$ ($-$ $\infty$, $-$3) $\cup$ [4, $\infty$) ...... (1)<br><br>Case II :<br><br>When ${\left| {[x]} \right| - 2}$ $\le$ 0<br><br>and ${\left| {[x]} \right| - 3}$ &lt; 0<br><br>$\therefore$ x $\in$ [$-$2, 3) ..... (2)<br><br>So, from (1) and (2) we get<br><br>Domain of function<br><br>= ($-$ $\infty$, $-$3) $\cup$ [$-$2, 3) $\cup$ [4, $\infty$)<br><br>$\therefore$ (a + b + c) = $-$3 + ($-$2) + 3 = $-$2 (a &lt; b &lt; c)<br><br>$\Rightarrow$ Option (3) is correct.