The equation of the circle, which cuts orthogonally each of the three circles x 2 + y 2 - 2 x + 3 y - 7 = 0 …

Assuming, we have a circle $x^2+y^2+2gx+2fy+c=0 ...\left(1\right)$ which cuts circles

 $x^2+y^2-2x+3y-7=0 ...\left(2\right)$, $x^2+y^2+5x-5y+9=0 ...\left(3\right)$ and $x^2+y^2+7x-9y+29=0 ...\left(4\right)$ orthogonally,

Applying condition for two circles intersecting orthogonally,

So, considering equations $\left(1\right)$ and $\left(2\right)$,

$2g\left(-1\right)+2f\left(\frac{3}{2}\right)=c-7\Rightarrow -2g+3f-c=-7 ...\left(4\right)$
Similarly, considering equations $\left(1\right)$ and $\left(3\right)$

$2g\left(\frac{5}{2}\right)+2f\left(-\frac{5}{2}\right)=c+9\Rightarrow 5g-5f-c=9 ...\left(5\right)$

and $\left(1\right)$ and $\left(4\right)$

$2g\left(\frac{7}{2}\right)+2f\left(-\frac{9}{2}\right)=c+29\Rightarrow 7g-9f-c=29 ...\left(6\right)$

Solving equations $\left(4\right)$, $\left(5\right)$ and $g=-8$, $f=-9$ and $c=-4$,

Hence, the equation of the circle is (From equation $\left(1\right)$)

$x^2+y^2-16x-18y-4=0$