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

The problem defines $$[x]$$ as the greatest integer function. The function under consideration is $$f(x) = \sqrt{\dfrac{|[x]|-2}{|[x]|-3}}$$, which requires $$\dfrac{|[x]|-2}{|[x]|-3} \ge 0$$ with $$|[x]| \ne 3$$.

Setting $$n = [x]$$ (an integer), the condition $$\dfrac{|n|-2}{|n|-3} \ge 0$$ holds when $$|n| \le 2$$ or $$|n| \ge 4$$ (excluding $$|n| = 3$$).

We translate each integer condition back to $$x$$, using $$[x] = n \iff x \in [n, n+1)$$.

For $$|n| \le 2$$: $$n \in \{-2,-1,0,1,2\}$$, so $$x \in [-2,-1) \cup [-1,0) \cup [0,1) \cup [1,2) \cup [2,3) = [-2, 3)$$.

For $$|n| = 3$$ (excluded): $$n = -3$$ gives $$x \in [-3,-2)$$, and $$n = 3$$ gives $$x \in [3,4)$$. These are excluded from the domain.

For $$|n| \ge 4$$: $$n \le -4$$ gives $$x \in (-\infty,-3)$$, and $$n \ge 4$$ gives $$x \in [4,\infty)$$.

Therefore the domain is $$(-\infty,-3) \cup [-2,3) \cup [4,\infty)$$.

Matching with the given form $$(-\infty, a) \cup [b,c) \cup [4,\infty)$$: we identify $$a = -3$$, $$b = -2$$, $$c = 3$$.

Hence $$a + b + c = -3 + (-2) + 3 = -2$$.