1. Find the set $S$

We need

$$S=\left\{x\in[-6,3]\setminus\{-2,2\}: \frac{|x+3|-1}{|x|-2}\ge 0\right\}.$$

So we solve

$$\frac{|x+3|-1}{|x|-2}\ge 0,$$

with domain restriction $x\in[-6,3]$ and $x\ne -2,2$.

Step 1: Simplify numerator

$$|x+3|-1\ge 0 \iff |x+3|\ge 1.$$

The numerator is zero when

$$|x+3|-1=0 \iff |x+3|=1 \iff x=-4,-2.$$

Also,

$$|x+3|-1<0 \iff |x+3|<1 \iff -4<x<-2.$$

Thus:

• Numerator $>0$ for $x<-4$ or $x>-2$
• Numerator $=0$ at $x=-4,-2$
• Numerator $<0$ for $-4<x<-2$

Step 2: Simplify denominator

$$|x|-2=0 \iff |x|=2 \iff x=\pm 2.$$

These are excluded already.

Now,

• $|x|-2>0$ when $|x|>2$, i.e. $x<-2$ or $x>2$
• $|x|-2<0$ when $|x|<2$, i.e. $-2<x<2$

Step 3: Sign analysis on $[-6,3]$

Critical points are

$$-4,-2,2.$$

Check intervals in $[-6,3]$:

(i) $[-6,-4)$

• Numerator $>0$
• Denominator $>0$

So fraction $>0$.

At $x=-4$, numerator $=0$, denominator $>0$, so fraction $=0$ and allowed.

Hence $[-6,-4] \subset S$.

(ii) $(-4,-2)$

• Numerator $<0$
• Denominator $>0$

So fraction $<0$, not allowed.

(iii) $(-2,2)$

• Numerator $>0$ (since here $x>-2$)
• Denominator $<0$

So fraction $<0$, not allowed.

(iv) $(2,3]$

• Numerator $>0$
• Denominator $>0$

So fraction $>0$, allowed.

Therefore,

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

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2. Find the set $T$

We need

$$T=\{x\in\mathbb Z: x^2-7|x|+9\le 0\}.$$

Let

$$y=|x|\ge 0.$$

Then the inequality becomes

$$y^2-7y+9\le 0.$$

Solve the quadratic:

$$y^2-7y+9=0.$$

Roots are

$$y=\frac{7\pm\sqrt{49-36}}{2}=\frac{7\pm\sqrt{13}}{2}.$$

So

$$\frac{7-\sqrt{13}}{2}\le y\le \frac{7+\sqrt{13}}{2}.$$

Now,

$$\frac{7-\sqrt{13}}{2}\approx 1.697, \qquad \frac{7+\sqrt{13}}{2}\approx 5.303.$$

Thus integer values of $y=|x|$ are

$$|x|=2,3,4,5.$$

Hence

$$T=\{-5,-4,-3,-2,2,3,4,5\}.$$

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3. Find $S\cap T$

We have

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

From $T$, the integers lying in $[-6,-4]$ are

$$-5,-4.$$

And the integers lying in $(2,3]$ are

$$3.$$

So

$$S\cap T=\{-5,-4,3\}.$$

Therefore, the number of elements is

$$3.$$

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4. Option check

• A: $7$ ❌
• B: $5$ ❌
• C: $4$ ❌
• D: $3$ ✅

So the correct option is D.