The given circles
Let $S_{1} \equiv x^{2}+y^{2}+6 x-1=0,$
Centre $C_{1}=(-3,0), \text { Radius } R_{1}=\sqrt{10}$
$S_{2} \equiv x^{2}+y^{2}-3 y+2=0,$
$C_{2}=(0,3 / 2), R_{2}=1 / 2$
$S_{3}=x^{2}+y^{2}+x+y-3=0$
$C_{3}=(-1 / 2,-1 / 2), R_{3}=\sqrt{7 / 2}$
Let $S \equiv x^{2}+y^{2}+2 g x+2 fy +c=0$ be equation of circle which cut orthogonal all three circles.
Then, by condition of orthogonality
$-2 g(-3)+2(-f)(0)=c-1$
$\Rightarrow 6 g=c-1$...(i)
$-2 g(0)+2 f(-3 / 2)=c+2$
$\Rightarrow -3 f=c+2$...(ii)
$-2 g(-1 / 2)+2 f(1 / 2)=c-3$
$\Rightarrow g +f=c-3$...(iii)
On solving Eqs. (i), (ii) and (iii), we get
$g=1 / 7, f=-9 / 7$
So, centre is $(-g,-f)=(-1 / 7,9 / 7)$