If $Z_1,Z_2$  are  complex numbers such that $\mathrm{Re}\left(Z_1\right)=|Z_1-1|,\mathrm{Re}\left(Z_2\right)=|Z_2-1|and\arg (Z_1-Z_2)=\frac{\pi }{6},then\mathrm{Im}(Z_1+Z_2)$
 is equal to :

(B) Let  $z_1=x_1+i{y}_{1}and{z}_2=x_2+i{y}_2$

$\begin{array}{l}\because |z_1-1|=\mathrm{Re}\left(z_1\right)\Rightarrow {(x_1-1)}^2+y_1^2=x_1^2\\ \Rightarrow y_1^2-2x_1+1=0\mathrm{.............}\left(i\right)\\ |z_2-1|=\mathrm{Re}\left(z_2\right)\Rightarrow {(x_2-1)}^2+y_2^2=x_2^2\\ \Rightarrow y_2^2-2x_2+1=0\mathrm{.............}\left(ii\right)\\ Fromeqn.\left(i\right)-\left(ii\right),\\ y_1^2-y_2^2-2(x_1-x_2)=0\Rightarrow y_1+y_2=2\left(\frac{x_1-x_2}{y_1-y_2}\right)\mathrm{.....}\left(iii\right)\\ \because \arg (z_1-z_2)=\frac{\pi }{6}\\ \Rightarrow {\tan }^{-1}\left(\frac{y_1-y_2}{x_1-x_2}\right)=\frac{\pi }{6}\Rightarrow \frac{y_1-y_2}{x_1-x_2}=\frac{1}{\sqrt{3}}\\ \Rightarrow \frac{2}{y_1-y_2}=\frac{1}{\sqrt{3}}[From,\frac{y_1-y_2}{x_1-x_2}=\frac{2}{y_1-y_2}]\\ \therefore y_1-y_2=2\sqrt{3}\Rightarrow lm(z_1-z_2)=2\sqrt{3}\end{array}$