If ${z_1} = 1 + i$,${z_2} = -2 + 3i$,and ${z_3} = \frac{ai}{3}$,where ${i^2} = -1$,are collinear,then the value of $a$ is:

The solutions of the equation in $z$,$|z|^2 - (z + \bar{z}) + i(z - \bar{z}) + 2 = 0$ are $(i = \sqrt{-1})$.

The vector $z = 3 - 4i$ is turned anticlockwise through an angle of $180^{\circ}$ and stretched $2.5$ times. The complex number corresponding to the newly obtained vector is

Let $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}$,where $z \in \mathbb{C}$,be the equation of a circle with center at $C$. If the area of the triangle,whose vertices are at the points $(0,0)$,$C$,and $(\alpha, 0)$,is $11$ square units,then $\alpha^2$ equals

If $z = x + iy$ and $|z - 2 + i| = |z - 3 - i|$,then the locus of $z$ is:

Let $S_1 = \{z \in \mathbb{C} : |z| \leq 5\}$, $S_2 = \{z \in \mathbb{C} : \operatorname{Im}\left(\frac{z+1-\sqrt{3}i}{1-\sqrt{3}i}\right) \geq 0\}$ and $S_3 = \{z \in \mathbb{C} : \operatorname{Re}(z) \geq 0\}$. Then the area of the region $S_1 \cap S_2 \cap S_3$ is: