Question 13
A cube with side length $10$ is suspended above a plane. The vertex closest to the plane is labeled $A.$ 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 $\frac{r - \sqrt{s}}{t},$ where $r,$ $s,$ and $t$ are positive integers. Find $r + s + t.$