The circle passing through the intersection of the circles, $x^2+y^2-6x=0$ and $x^2+y^2-4y=0$ having its centre on the line, $2x-3y+12=0$, also passes through the point :

By family of circle, circle passing through intersection of given circles will be member of

${\mathrm{S}}_1+{\mathrm{\lambda S}}_2=0\text{ family }(\mathrm{\lambda }\ne 1)$

$\left({\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}\right)+\mathrm{\lambda }\left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}\right)=0$

$(\mathrm{\lambda }+1){\mathrm{x}}^2+(\mathrm{\lambda }+1){\mathrm{y}}^2-6\mathrm{x}-4\mathrm{\lambda y}=0$

${\mathrm{x}}^2+{\mathrm{y}}^2-\frac{6}{\mathrm{\lambda }+1}\mathrm{x}-\frac{4\mathrm{\lambda }}{\mathrm{\lambda }+1}\mathrm{y}=0$

$\text{ Centre }\left(\frac{3}{\lambda +1},\frac{2\lambda }{\lambda +1}\right)$

Centre lies on $2\mathrm{x}-3\mathrm{y}+12=0$

$2\left(\frac{3}{\mathrm{\lambda }+1}\right)-3\left(\frac{2\mathrm{\lambda }}{\mathrm{\lambda }+1}\right)+12=0$

$6\mathrm{\lambda }+18=0$

$\mathrm{\lambda }=-3$

Now circle is:${\mathrm{x}}^2+{\mathrm{y}}^2+3\mathrm{x}-6\mathrm{y}=0$

Now check each option