The value of $|z|^2+|z-3|^2+|z-i|^2$ is minimum when $z$ equals

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:
$(A) -6 \text{ and } 3, (B) -3 \text{ and } 6, (C) -6 \text{ and } 6, (D) -3 \text{ and } 9$

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a

The locus of $z$ satisfying $|z|+|z-1|=3$ is

Let $a, b \in \mathbb{R}$ and $a^2+b^2 \neq 0$. Suppose $S = \{z \in \mathbb{C} : z = \frac{1}{a+ibt}, t \in \mathbb{R}, t \neq 0\}$,where $i = \sqrt{-1}$. If $z = x+iy$ and $z \in S$,then $(x, y)$ lies on:

The locus of $z$ satisfying $\left|\frac{z-i}{z-2i}\right|=2$ is a