If one of the diameters of the circle, given by the equation $x^2+y^2-4 x+6 y-12=0$, is a chord of a circle,…

Find the radius of circle $S$ given its center at $(-3, 2)$ and that a diameter of $C_1$ is a chord of $S$.

For circle $C_1$ defined by $x^2+y^2-4x+6y-12=0$, comparing with the general form $x^2+y^2+2gx+2fy+c=0$ gives $g = -2$, $f = 3$, and $c = -12$.

The center of $C_1$ is $(2, -3)$ and its radius is $r_1 = \sqrt{g^2+f^2-c} = \sqrt{4+9+12} = 5$.

The diameter of $C_1$ has length $10$ and serves as a chord of $S$. The perpendicular distance from the center of $S$ at $(-3, 2)$ to this chord is the distance to the midpoint $(2, -3)$: $d = \sqrt{(2+3)^2+(-3-2)^2} = \sqrt{25+25} = 5\sqrt{2}$.

Using the right triangle formed by the center of $S$, the chord midpoint, and an endpoint, apply the Pythagorean theorem: $R^2 = d^2 + \left(\frac{10}{2}\right)^2 = (5\sqrt{2})^2 + 5^2 = 50 + 25 = 75$.

$R = \sqrt{75} = 5\sqrt{3}$

Final answer: $\boxed{C}$