Number of Elements in Intersection of Complex Sets

To find the number of elements in the set $A \cap B \cap C$, let $z = x + iy$, where $x, y \in \mathbb{R}$.

1. Geometric interpretation of the given sets:

• Set $A$:
  $A = \{z : \operatorname{Im} z \ge 1\} \implies y \ge 1$
  This represents the half-plane on and above the horizontal line $y = 1$.

• Set $B$:
  $B = \{z : |z - (2 + i)| = 3\}$
  This represents a circle with center $P(2, 1)$ and radius $R = 3$:
  $(x - 2)^2 + (y - 1)^2 = 9$

• Set $C$:
  $(1 - i)z = (1 - i)(x + iy) = (x + y) + i(y - x)$
  Therefore,
  $\operatorname{Re}((1 - i)z) = x + y$
  Hence,
  $C = \{z : x + y = \sqrt{2}\}$
  This represents the straight line $x + y = \sqrt{2}$.

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2. Intersection of $B$ and $C$:

The distance $d$ from the center of the circle $P(2, 1)$ to the line $x + y - \sqrt{2} = 0$ is:
$d = \frac{|2 + 1 - \sqrt{2}|}{\sqrt{1^2 + 1^2}} = \frac{3 - \sqrt{2}}{\sqrt{2}} = \frac{3}{\sqrt{2}} - 1 \approx 1.121$

Since $d < R = 3$, the line $C$ intersects the circle $B$ at two distinct points.

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3. Intersection with set $A$ ($y \ge 1$):

From the equation of the line $C$, we have $x = \sqrt{2} - y$. Substituting this into the circle equation:
$((\sqrt{2} - y) - 2)^2 + (y - 1)^2 = 9$
$(y + 2 - \sqrt{2})^2 + (y - 1)^2 = 9$

Expanding and simplifying the equation:
$\left[y^2 + 2(2 - \sqrt{2})y + (2 - \sqrt{2})^2\right] + \left[y^2 - 2y + 1\right] = 9$
$2y^2 + (2 - 2\sqrt{2})y + (6 - 4\sqrt{2}) + 1 - 9 = 0$
$2y^2 + 2(1 - \sqrt{2})y - 2(1 + 2\sqrt{2}) = 0$
$y^2 + (1 - \sqrt{2})y - (1 + 2\sqrt{2}) = 0$

Let $f(y) = y^2 + (1 - \sqrt{2})y - (1 + 2\sqrt{2})$. Evaluating $f(y)$ at $y = 1$:
$f(1) = 1^2 + (1 - \sqrt{2})(1) - (1 + 2\sqrt{2}) = 1 - 3\sqrt{2} < 0$

Since the parabola $f(y)$ opens upwards and $f(1) < 0$, the two real roots $y_1$ and $y_2$ (with $y_1 < y_2$) satisfy:
$y_1 < 1 < y_2$

Thus, exactly one of the two intersection points has $y \ge 1$.

Hence, the number of elements in the set $A \cap B \cap C$ is $1$.

Therefore, the correct option is B.