A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\tfrac{r-\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers, and $r+s+t<1000$. Find $r+s+t$.