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…

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 $5\sqrt{3}$