Let, $Z=x+iy$
Now,
$\left(Z-2\right)\left(\overline{Z}-1\right)=Z\overline{Z}-2\overline{Z}-Z+2$
$=x^2+y^2-3x+2+iy$
argument of $\left(1-i\right)\left(\sqrt{3}+i\right)\left(1+\sqrt{3}\mathrm{}i\right)=-\frac{\pi }{4}+\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{4}$

Now, since both arguments are equal
$\Rightarrow$ $x^2+y^2-3x+2=y (\mathrm{where}, y>0)$
$\Rightarrow$ $x^2+y^2-3x-y+2=0$
Hence, the centre of the circle is $\left(\frac{3}{2},\frac{1}{2}\right)=\left(a,b\right)$
$\Rightarrow$ $\frac{1}{a+b}=\frac{1}{\frac{3}{2}+\frac{1}{2}}=\frac{1}{2}=0.5$