The farthest point on the circle $x^2+y^2-4x+6y-12=0$ from $\left(-13, 17\right)$ is 

$\left(5, -7\right) and \left(-1, 1\right)$ are points on the given circle.

Distance between (5, -7) and (-13, 17) is $\sqrt{{18}^2+{24}^2}=\sqrt{324+576}=\sqrt{900}=30$

Distance between $\left(-1, 1\right) and \left(-13, 17\right) is \sqrt{{12}^2+{16}^2}=\sqrt{144+256}=\sqrt{400}= 20$

$\therefore$ $\left(5, -7\right)$ is the farthest point.