jmc

As the center of the circle is at the point $(2,3)$, and one point on the circle is at the point $(-1,6)$, by the distance formula, the radius of the circle is $\sqrt{(2-(-1){)}^2+(3-6{)}^2}=\sqrt{3^2+3^2}=\sqrt{18}$. The equation of the circle is then given by $(x-2{)}^2+(y-3{)}^2=18$, and expanding, $$x^2-4x+4+y^2-6y+9-18=0⟹x^2+y^2-4x-6y-5=0.$$ Thus, $A\times B\times C=-4\times -6\times -5=\boxed{-120}$.