Let $z=x+iy$, where $x,y\in \mathbb{R}$.

We interpret each set geometrically.

1. Set $S_1$

Given $|z-3-2i|^2=8$ This means $|(x-3)+i(y-2)|^2=(x-3)^2+(y-2)^2=8$ So $S_1$ is the circle with center $(3,2)$ and radius $2\sqrt{2}$.

2. Set $S_2$

Given $\operatorname{Re}(z)\ge 5$ Since $\operatorname{Re}(z)=x$, we get $x\ge 5$ So $S_2$ is the half-plane to the right of the vertical line $x=5$.

3. Set $S_3$

Given $|z-\overline z|\ge 8$ Now, $z-\overline z=(x+iy)-(x-iy)=2iy$ Hence $|z-\overline z|=|2iy|=2|y|$ So $2|y|\ge 8 \implies |y|\ge 4$ Thus $S_3$ is the region where $y\ge 4 \quad \text{or} \quad y\le -4$

4. Find $S_1\cap S_2\cap S_3$

We need points on the circle $(x-3)^2+(y-2)^2=8$ with $x\ge 5, \qquad |y|\ge 4$

Step 1: Use $x\ge 5$

Since the circle has center $x=3$ and radius $2\sqrt2$, its rightmost point is $x=3+2\sqrt2$ Since $2\sqrt2\approx 2.828$, this is about $5.828$, so points with $x\ge 5$ are possible.

Put $x=5$ in the circle equation: $(5-3)^2+(y-2)^2=8$ $4+(y-2)^2=8$ $(y-2)^2=4$ $y=4 \text{ or } y=0$ Thus the line $x=5$ meets the circle at $(5,4)$ and $(5,0)$.

For $x>5$, the relevant arc lies between these two points, so along this arc we have $0<y<4$ except at the endpoint $(5,4)$.

Step 2: Apply $|y|\ge 4$

From the circle equation, the maximum and minimum possible values of $y$ are $y=2\pm 2\sqrt2$ So

Hence on the circle, $y\le -4$ is impossible.

So only $y\ge 4$ can work. Now solve simultaneously: $(x-3)^2+(y-2)^2=8, \qquad x\ge 5, \qquad y\ge 4$ At $y=4$, $(x-3)^2+(4-2)^2=8$ $(x-3)^2+4=8$ $(x-3)^2=4$ $x=5 \text{ or } x=1$ Among these, only $x=5$ satisfies $x\ge 5$. So the only common point is $z=5+4i$

Therefore, $|S_1\cap S_2\cap S_3|=1$

5. Compare with stored answer

Our derived answer is $1$, which matches option A.