$\text{ The circle passing through the intersection of the circles, }{\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}=0\text{ and }{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}=0\text{ , }$

$\text{ having its centre on the line, }2\mathrm{x}-3\mathrm{y}+12=0,\text{ also passes through the point: }$

$\begin{array}{l}{x}^2+y^2-6x=0\\ x^2+y^2-4y=0\end{array}$

$Equation of any circle pas\sin g through the points of intersection of the above two circles is$

$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}+\lambda \left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}\right)=0\\ \text{ Its centre }\left(\frac{3}{1+\lambda },\frac{2\lambda }{1+\lambda }\right)\end{array}$

$\begin{array}{l}\text{ It lies on }2x-3y+12=0 \Rightarrow \frac{6}{1+\lambda }-\frac{6\lambda }{1+\lambda }+12=0\\ 6-6\lambda +12+12\lambda =0, 6\lambda =-18\\ \lambda =-3\\ {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-3\left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{y}\right)=0\\ -2x^2-2y^2-6x+12y=0\\ x^2+y^2+3x-6y=0\\ \text{ It passes through (by substituting the options ), required point is }(-3,6)\\ \end{array}$