The value of $\frac{z_1-z_2}{z_3}-1$ is

Further $\arg ({\mathrm{z}}_1-{\mathrm{z}}_2)=\frac{\pi }{3}$ as $OABC$ is a rhombus all sides are equal.
i.e $|Z_1|=|Z_3|=|Z_1-Z_2|=2$  as $Z_1$ and $Z_3$ lie on $|Z|=2$
According to parallelogram law of edition $Z_1+Z_3=Z_2$
$\Rightarrow Z_1-Z_2=-Z_3$ ------(1)
$\therefore \frac{Z_1-Z_2}{Z_3}-1=-1-1=-2$    (from equation (1))
Hence, option A.