The circle passing through the intersection of

Let $S$ be the circle pasing through point of intersection of $\mathrm{S}_{1} \& \mathrm{~S}_{2}$

$\therefore \quad \mathrm{S}=\mathrm{S}_{1}+\lambda \mathrm{S}_{2}=0$

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

$\Rightarrow S: x^{2}+y^{2}-\left(\frac{6}{1+\lambda}\right) x-\left(\frac{4 \lambda}{1+\lambda}\right) y=0$

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

$2 x-3 y+12=0 \Rightarrow \lambda=-3$

put in $(1) \Rightarrow S: x^{2}+y^{2}+3 x-6 y=0$

Now check options point $(-3,6)$

lies on $\mathrm{S}$.