Let $$S = \{x \in [-6, 3] - \{-2, 2\} : \frac{|x+3|-1}{|x|-2} \geq 0\}$$ and $$T = \{x \in \mathbb{Z} : x^2 - 7|x| + 9 \leq 0\}$$. Then the number of elements in $$S \cap T$$ is

We need to find the sets $$S$$ and $$T$$ and then determine the number of elements in $$S \cap T$$.

We have $$S = \{x \in [-6, 3] \setminus \{-2, 2\} : \frac{|x+3|-1}{|x|-2} \geq 0\}$$.

The numerator $$|x+3| - 1 \geq 0$$ when $$|x+3| \geq 1$$, i.e., $$x \leq -4$$ or $$x \geq -2$$. The numerator is negative when $$-4 < x < -2$$.

The denominator $$|x| - 2 > 0$$ when $$x < -2$$ or $$x > 2$$, and $$|x| - 2 < 0$$ when $$-2 < x < 2$$. The points $$x = \pm 2$$ are excluded.

For the fraction to be non-negative, we need both numerator and denominator to share the same sign (or numerator equals zero).

Case 1: Numerator $$\geq 0$$ and denominator $$> 0$$. This requires $$(x \leq -4 \text{ or } x \geq -2)$$ and $$(x < -2 \text{ or } x > 2)$$. Within $$[-6, 3] \setminus \{-2, 2\}$$, this gives $$[-6, -4] \cup (2, 3]$$.

Case 2: Numerator $$\leq 0$$ and denominator $$< 0$$. This requires $$-4 < x < -2$$ and $$-2 < x < 2$$. These intervals do not overlap, so this case contributes nothing.

Therefore $$S = [-6, -4] \cup (2, 3]$$.

Now for $$T$$, we solve $$x^2 - 7|x| + 9 \leq 0$$ over integers. Setting $$u = |x|$$, we get $$u^2 - 7u + 9 \leq 0$$. The roots are $$u = \frac{7 \pm \sqrt{13}}{2}$$, which gives approximately $$u \in [1.697, 5.303]$$. So for integer values of $$u$$: $$u \in \{2, 3, 4, 5\}$$, and therefore $$T = \{-5, -4, -3, -2, 2, 3, 4, 5\}$$.

Now $$S \cap T$$ consists of elements of $$T$$ that lie in $$S = [-6, -4] \cup (2, 3]$$. From $$[-6, -4]$$, we get $$\{-5, -4\}$$. From $$(2, 3]$$, we get $$\{3\}$$. So $$S \cap T = \{-5, -4, 3\}$$, which has 3 elements.

Hence, the correct answer is Option 4.