As $A$ and $B$ satisfy both line and circle we have $\alpha=0 \Rightarrow A(0,0)$ and $\beta=1$ i.e. $B(1,1)$

Centre of circle as $A B$ diameter is $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $=\frac{1}{\sqrt{2}}$

$\therefore$ For image of $\left(\frac{1}{2} ; \frac{1}{2}\right)$ in $x+y+z$ we get $\frac{x-\frac{1}{2}}{1}=\frac{y-\frac{1}{2}}{1}=\frac{-2(3)}{2}$

$\Rightarrow$ Image $\left(-\frac{5}{2},-\frac{5}{2}\right)$

$\therefore $ Equation of required circle

$\left(x+\frac{5}{2}\right)^{2}+\left(y+\frac{5}{2}\right)^{2}=\frac{1}{2}$

$\Rightarrow x^{2}+y^{2}+5 x+5 y+\frac{50}{4}-\frac{1}{2}=0$

$\Rightarrow x^{2}+y^{2}+5 x+5 y+12=0$