Step-by-step Solution:

The problem asks for the number of elements in the intersection of three sets of complex numbers A, B, and C. Let the complex number z be represented as z = x + iy, where x and y are real numbers.

1. Analyze Set A: A = {z : Im(z) >= 1} Since z = x + iy, Im(z) = y. The condition for set A is y >= 1. Geometrically, this represents the closed upper half-plane including the line y = 1.

2. Analyze Set B: B = {z : |z - 2 - i| = 3} This can be written as |z - (2 + i)| = 3. This equation represents all complex numbers z whose distance from the complex number 2 + i is 3. Geometrically, this is a circle with center (2, 1) and radius r = 3. The Cartesian equation of this circle is $(x - 2)^2 + (y - 1)^2 = 3^2 = 9$.

3. Analyze Set C: C = {z : Re((1 - i)z) = sqrt(2)} Let's compute (1 - i)z: $(1 - i)(x + iy) = x + iy - ix - i^2y = x + y + i(y - x)$ The real part of this is Re((1 - i)z) = x + y. The condition for set C is x + y = sqrt(2). Geometrically, this is a straight line.

4. Find the intersection A ∩ B ∩ C: We need to find the number of points (x, y) that satisfy all three conditions simultaneously: (i) y >= 1 (ii) $(x - 2)^2 + (y - 1)^2 = 9$ (iii) x + y = sqrt(2)

First, let's find the intersection of the circle (B) and the line (C). We can do this by substituting the expression for x from the line equation into the circle equation. From (iii), x = sqrt(2) - y. Substitute this into (ii): $( (sqrt(2) - y) - 2 )^2 + (y - 1)^2 = 9$ $( (sqrt(2) - 2) - y )^2 + (y - 1)^2 = 9$ $(y - (sqrt(2) - 2))^2 + (y - 1)^2 = 9$ Expanding the terms: $[y^2 - 2(sqrt(2) - 2)y + (sqrt(2) - 2)^2] + [y^2 - 2y + 1] = 9$ $y^2 - (2sqrt(2) - 4)y + (2 - 4sqrt(2) + 4) + y^2 - 2y + 1 = 9$ $2y^2 + (-2sqrt(2) + 4 - 2)y + (6 - 4sqrt(2) + 1) = 9$ $2y^2 + (2 - 2sqrt(2))y + (7 - 4sqrt(2)) = 9$ $2y^2 - 2(sqrt(2) - 1)y - (2 + 4sqrt(2)) = 0$ Dividing by 2: $y^2 - (sqrt(2) - 1)y - (1 + 2sqrt(2)) = 0$

This is a quadratic equation for y, which represents the y-coordinates of the intersection points of the circle B and the line C. A quadratic equation can have 0, 1, or 2 real roots. The discriminant $Δ = b^2 - 4ac$ will tell us the number of intersection points. a = 1, b = -(sqrt(2) - 1), c = -(1 + 2sqrt(2)) $Δ = (-(sqrt(2) - 1))^2 - 4(1)(-(1 + 2sqrt(2)))$ $Δ = (sqrt(2) - 1)^2 + 4(1 + 2sqrt(2))$ Δ = (2 - 2sqrt(2) + 1) + (4 + 8sqrt(2)) = 3 - 2sqrt(2) + 4 + 8sqrt(2) = 7 + 6sqrt(2) Since Δ = 7 + 6sqrt(2) > 0, there are two distinct real roots for y. This means the line C intersects the circle B at two distinct points.

5. Check condition for Set A: Now we must check how many of these intersection points satisfy the condition for set A, which is y >= 1. Let the roots of the quadratic equation $f(y) = y^2 - (sqrt(2) - 1)y - (1 + 2sqrt(2)) = 0$ be $y_1$ and $y_2$. We can determine the position of the roots relative to y=1 by evaluating the function f(1). $f(1) = (1)^2 - (sqrt(2) - 1)(1) - (1 + 2sqrt(2))$ f(1) = 1 - sqrt(2) + 1 - 1 - 2sqrt(2) f(1) = 1 - 3sqrt(2) Since 3sqrt(2) = sqrt(18) and 1 = sqrt(1), we have 3sqrt(2) > 1. Therefore, 1 - 3sqrt(2) < 0.

The quadratic function f(y) represents an upward-opening parabola (since the coefficient of $y^2$ is positive). Since f(1) < 0, the value y = 1 must lie between the two roots $y_1$ and $y_2$. So, one root is less than 1 and the other is greater than 1. Let's say $y_1 < 1 < y_2$.

This means that of the two intersection points of B and C:

• One point has a y-coordinate $y_2 > 1$, which satisfies the condition y >= 1. This point is in set A.
• The other point has a y-coordinate $y_1 < 1$, which does not satisfy the condition y >= 1. This point is not in set A.

Therefore, there is exactly one point that lies in all three sets A, B, and C. The number of elements in the set A ∩ B ∩ C is 1.

Final Answer is 1.