The point A, B, C distinct complex numbers represented by $Z_1,Z_2,Z_3$ respectively. On a complex plane and angle B and C of the triangle ABC are each equal to $\frac{\pi }{2}-\frac{\alpha }{2}$ . If ${\left(Z_2–Z_3\right)}^2=\lambda \left(Z_3–Z_1\right)\left(Z_1–Z_2\right){\sin }^2\frac{\alpha }{2}$.
 Then find the value of ‘$\lambda$ ’. 

$$\begin{gathered} {\left({\mathrm{Z}}_2-{\mathrm{Z}}_3\right)}^2=\lambda \left({\mathrm{Z}}_3-{\mathrm{Z}}_1\right)\left({\mathrm{Z}}_1-{\mathrm{Z}}_2\right){\sin }^2\frac{\alpha }{2}\dots \dots \dots \dots \left(1\right) \\ \text{ Now }{\mathrm{Z}}_3-{\mathrm{Z}}_1=\left(Z_2-Z_1\right){\mathrm{e}}^{\mathrm{i}\alpha } ...........\left(2\right) \end{gathered}$$

$$\begin{gathered} \frac{z_3-z_1}{z_2-z_1}-1=\cos \alpha +i\sin \alpha -1 \\ \Rightarrow \frac{z_3-z_2}{z_2-z_1}=-2{\sin }^2\frac{\alpha }{2}\text{+}2i\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}=i2\sin \frac{\alpha }{2}\left[\cos \frac{\alpha }{2}+i\sin \frac{\alpha }{2}\right] \\ \Rightarrow {\left(\frac{z_2-z_3}{z_1-z_2}\right)}^2=-4{\sin }^2\frac{\alpha }{2}{e}^{i\alpha } \\ \Rightarrow =-4{\sin }^2\frac{\alpha }{2}\left(\frac{z_3-z_1}{z_2-z_1}\right)from\left(2\right) \\ \Rightarrow \frac{{(z_2-z_3)}^2}{{(z_1-z_2)}^2}=4{\sin }^2\frac{\alpha }{2}\left(\frac{z_3-z_1}{z_1-z_2}\right) \\ \Rightarrow \frac{{(z_2-z_3)}^2}{(z_1-z_2)}=4{\sin }^2\frac{\alpha }{2}(z_3-z_1) \\ \Rightarrow {(z_2-z_3)}^2=4(z_3-z_1)(z_1-z_2){\sin }^2\frac{\alpha }{2} So \lambda = 4 \end{gathered}$$