Let S = \ x ∈ [- 6, 3] - \ - 2, 2\ : (|x + 3| - 1)/(|x| - 2) ≥ 0\...

Given, ${\left|x\right|}^{2} - 7\left|x\right| + 9 \le 0$

$⇒ \left|x\right| \le \dfrac{7 \pm \sqrt{13}}{2}$

$⇒ \left|x\right| \in \left[\dfrac{7 - \sqrt{13}}{2}, \dfrac{7 + \sqrt{13}}{2}\right]$

Also given $x \in \text{Integers}$, so $x$ can be $\pm 2, \pm 3, \pm 4, \pm 5$

Now out of these values of $x$ only $3, - 4, - 5$ will satisfy $S = \dfrac{\left|x + 3\right| - 1}{\left|x\right| - 2} \ge 0$

So, $S \cap T \in \left\{3, - 4, - 5\right\}$

So, $n\left(S \cap T\right) = 3$