The equation of polar w.r.t the point $(1,1)$ to the circle $x^2+y^2-4 x-6 y+12=0$ is
$$
\begin{array}{rlrl}
x \cdot 1+y \cdot 1-2(x+1)-3(y+1)+12 & =0 \\
\Rightarrow & -x-2 y+7 & =0 \\
\Rightarrow & x+2 y-7 & =0
\end{array}
$$
Since, the inverse of the point $(1,1)$ is the foot $(h, k)$ of the perpendicular from the point $(1,1)$ to the line $x+2 y-7=0$
So,
$$
\begin{aligned}
\text { So, } & & \frac{h-1}{1} & =\frac{k-1}{2}=\frac{-(1+2-7)}{(1)^2+(2)^2} \\
\Rightarrow & & h-1 & =\frac{4}{5} \Rightarrow h=\frac{9}{5} \\
& \text { and } & \frac{k-1}{2} & =\frac{4}{5} \Rightarrow k-1=\frac{8}{5} \\
\Rightarrow & & k & =13 / 5 \\
& \therefore & h+k & =\frac{9}{5}+\frac{13}{5}=\frac{22}{5}
\end{aligned}
$$