Let $$C$$ be the set of all complex numbers. Let
$$S_1 = \{z \in C \mid |z - 3 - 2i|^2 = 8\}$$,
$$S_2 = \{z \in C \mid \text{Re}(z) \geq 5\}$$ and
$$S_3 = \{z \in C \mid |z - \bar{z}| \geq 8\}$$.
Then the number of elements in $$S_1 \cap S_2 \cap S_3$$ is equal to

Let a general complex number be written as $$z = x + iy$$ where $$x = \text{Re}(z)$$ and $$y = \text{Im}(z)$$. We now translate the three set-descriptions into algebraic conditions in $$x$$ and $$y$$ so that we can see their mutual intersection clearly.

From $$S_1$$ we have

$$|z - (3 + 2i)|^2 = 8.$$

Using the fact that $$|a + ib|^2 = a^2 + b^2$$, we write

$$|(x+iy) - (3 + 2i)|^2 = |(x-3) + i(y-2)|^2 = (x-3)^2 + (y-2)^2.$$

So the condition for $$S_1$$ becomes

$$ (x-3)^2 + (y-2)^2 = 8. \quad -(1)$$

This is the equation of a circle with centre $$(3,\,2)$$ and radius $$\sqrt{8} = 2\sqrt2 \;(\approx 2.828).$$

From $$S_2$$ we are told that the real part is at least $$5$$, so

$$\text{Re}(z) \ge 5 \quad\Longrightarrow\quad x \ge 5. \quad -(2)$$

For $$S_3$$ we use the conjugate. Because $$\bar z = x - iy$$, we have

$$z - \bar z = (x + iy) - (x - iy) = 2iy.$$

Now, $$|2iy| = 2|y|,$$ therefore the condition $$|z - \bar z|\ge 8$$ gives

$$2|y| \ge 8 \quad\Longrightarrow\quad |y| \ge 4 \quad\Longrightarrow\quad y \ge 4 \text{ or } y \le -4. \quad -(3)$$

We now want the simultaneous satisfaction of (1), (2) and (3). In other words, we need all points on the circle (1) that also lie in the half-plane $$x\ge 5$$ and simultaneously have either $$y\ge 4$$ or $$y\le -4$$.

Because the circle’s centre is $$(3,2)$$ and its radius is $$2\sqrt2\approx 2.828,$$ its extreme $$x$$-coordinates are

$$x_{\min}=3-2\sqrt2\approx 0.172,\qquad x_{\max}=3+2\sqrt2\approx5.828.$$

Condition (2) restricts us to

$$5 \le x \le 5.828. \quad -(4)$$

Let us keep $$x$$ within the interval (4) and solve (1) for $$y$$. Starting from (1), we isolate the $$y$$-part:

$$ (y-2)^2 = 8 - (x-3)^2. $$

Taking the square root of both sides gives

$$ y - 2 = \pm\sqrt{\,8 - (x-3)^2\,}. \quad -(5)$$

Define

$$t = \sqrt{\,8 - (x-3)^2\,}, \qquad t\ge 0.$$

Then the two possible $$y$$ values are

$$ y = 2 + t \quad\text{or}\quad y = 2 - t. \quad -(6)$$

We next impose (3). First consider $$y \ge 4.$$ Using (6) we have

$$2 + t \ge 4 \quad\Longrightarrow\quad t \ge 2. \quad -(7)$$

Because $$t=\sqrt{\,8 - (x-3)^2\,},$$ inequality (7) squares to

$$8 - (x-3)^2 \ge 4 \quad\Longrightarrow\quad (x-3)^2 \le 4. \quad -(8)$$

Inequality (8) is the same as

$$-2 \le x - 3 \le 2 \quad\Longrightarrow\quad 1 \le x \le 5. \quad -(9)$$

Combining (9) with the earlier restriction (4), we must have

$$5 \le x \le 5.828\quad\text{and}\quad 1 \le x \le 5.$$

The only common value is

$$x = 5.$$

Substituting $$x=5$$ back into (1) to find $$y$$, we compute

$$(x-3)^2 = (5-3)^2 = 4,$$

and hence

$$(y-2)^2 = 8 - 4 = 4.$$

Therefore

$$y - 2 = \pm 2 \quad\Longrightarrow\quad y = 4 \ \text{or}\ y = 0.$$

From (3) we need $$|y| \ge 4,$$ so $$y = 0$$ is discarded while $$y = 4$$ is accepted.

We must also test the case $$y \le -4$$. However, the circle’s lowest $$y$$-coordinate is the centre coordinate minus the radius, namely $$2 - 2\sqrt2 \approx -0.828,$$ which is already greater than $$-4.$$ Thus no point on the circle can satisfy $$y\le -4.$$

Consequently, exactly one point survives all three conditions, namely

$$z = 5 + 4i.$$

There is only one such complex number, so the intersection $$S_1 \cap S_2 \cap S_3$$ contains exactly one element.

Hence, the correct answer is Option A.