If three complex numbers are in $A.P.$,then they lie on

If $z_{1}, z_{2}$ are complex numbers such that $\operatorname{Re}(z_{1})=|z_{1}-1|$, $\operatorname{Re}(z_{2})=|z_{2}-1|$ and $\arg(z_{1}-z_{2})=\frac{\pi}{6}$, then $\operatorname{Im}(z_{1}+z_{2})$ is equal to

Let $z_1 = 6 + i$ and $z_2 = 4 - 3i$. Let $z$ be a complex number such that $\arg \left( \frac{z - z_1}{z_2 - z} \right) = \frac{\pi}{2}$,then $z$ satisfies -

If complex numbers ${z_1}, {z_2}, \text{and } {z_3}$ represent the vertices $A, B, \text{and } C$ respectively of an isosceles triangle $ABC$ of which $\angle C$ is a right angle,then the correct statement is:

Let $S = \{z \in \mathbb{C} - \{i, 2i\} : \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R} \}$. If $\alpha - \frac{13}{11}i \in S$ and $\alpha \in \mathbb{R} - \{0\}$,then $242\alpha^2$ is equal to

If $|z-25i| \leq 15$,then the value of $\text{Maximum } \arg(z) - \text{Minimum } \arg(z)$ is equal to