If one of the diameters of the circle, given by the equation, x ^ 2...

The centre of the given circle ${x}^{2} + {y}^{2} - 4x + 6y - 12 = 0$ is $\left(2, - 3\right)$ and the radius is $5$.

The distance between the centres $5\sqrt{2}$ and radius is $5$. The triangle $OPQ$ is a right angled triangle

$OQ = \sqrt{{\left(5\sqrt{2}\right)}^{2} + {5}^{2}} = \sqrt{{\left(5\sqrt{3}\right)}^{2}} = 5\sqrt{3}$

Hence answer is $5\sqrt{3}$