The complex numbers $z_1,z_{2}and{z}_3$ satisfying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$ are the vertices of a triangle which is

Taking modulus on both sides of given relation. 

$\frac{\left|z_1-z_3\right|}{\left|z_2-z_3\right|}=\frac{1}{4}+\frac{3}{4}=1\therefore AC=BC$

Hence, $\mathrm{\Delta }$ is isosceles.

$\frac{z_1-z_3}{z_2-z_2}=\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}=e^{\frac{\pi }{3}i}$

Or $\left(z_2-z_3\right)=\left(z_1-z_3\right)e^{\frac{\pi }{3}i}$

Anticlockwise rotation implies that $\mathrm{\angle }ABC={60}^{\circ }\text{. }$ Hence triangle is equilateral.