Let $A, B, C$ be three sets of complex numbers as defined below:
$A = \{z : \operatorname{Im}(z) \geq 1\}$
$B = \{z : |z - 2 - i| = 3\}$
$C = \{z : \operatorname{Re}((1 - i)z) = \sqrt{2}\}$
$1.$ The number of elements in the set $A \cap B \cap C$ is:
$(A) 0, (B) 1, (C) 2, (D) \infty$
$2.$ Let $z$ be any point in $A \cap B \cap C$. Then,$|z + 1 - i|^2 + |z - 5 - i|^2$ lies between:
$(A) 25 \text{ and } 29, (B) 30 \text{ and } 34, (C) 35 \text{ and } 39, (D) 40 \text{ and } 44$
$3.$ Let $z$ be any point in $A \cap B \cap C$ and let $w$ be any point satisfying $|w - 2 - i| < 3$. Then,$|z| - |w| + 3$ lies between:
$(A) -6 \text{ and } 3, (B) -3 \text{ and } 6, (C) -6 \text{ and } 6, (D) -3 \text{ and } 9$

(B, C, D) $1.$ $A$ represents the region $y \geq 1$. $B$ is a circle with center $(2, 1)$ and radius $3$. $C$ is the line $\operatorname{Re}((1-i)(x+iy)) = x+y = \sqrt{2}$.
Substituting $y = \sqrt{2}-x$ into the circle equation $(x-2)^2 + (y-1)^2 = 9$ gives $(x-2)^2 + (\sqrt{2}-x-1)^2 = 9$.
Expanding this: $(x^2 - 4x + 4) + (x^2 + 2x(1-\sqrt{2}) + (1-\sqrt{2})^2) = 9$.
$2x^2 + x(2-2\sqrt{2}-4) + 4 + 1 - 2\sqrt{2} + 2 = 9 \implies 2x^2 - (2+2\sqrt{2})x - 2 - 2\sqrt{2} = 0$.
Solving this quadratic,we find one point with $y \geq 1$. Thus,the number of elements is $1$.
$2.$ Let $z = x+iy$. The expression is $|(x+1)+i(y-1)|^2 + |(x-5)+i(y-1)|^2 = (x+1)^2 + (y-1)^2 + (x-5)^2 + (y-1)^2$.
Since $z$ is on the circle $(x-2)^2 + (y-1)^2 = 9$,we have $(y-1)^2 = 9 - (x-2)^2$.
Substituting this,the expression becomes $(x+1)^2 + (x-5)^2 + 2(9 - (x-2)^2) = x^2+2x+1 + x^2-10x+25 + 18 - 2(x^2-4x+4) = 2x^2-8x+26 + 18 - 2x^2+8x-8 = 36$.
Since $36$ is between $35$ and $39$,the answer is $(C)$.
$3.$ Since $|z-2-i|=3$ and $|w-2-i| < 3$,by the triangle inequality $|z-w| < |z-(2+i)| + |w-(2+i)| < 3+3 = 6$.
Also $|z-w| > ||z-(2+i)| - |w-(2+i)|| = |3 - |w-(2+i)|| > 0$.
Thus $0 < |z-w| < 6$. Using $||z|-|w|| \leq |z-w|$,we have $-6 < |z|-|w| < 6$.
Adding $3$,we get $-3 < |z|-|w|+3 < 9$.