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 $r+s+t$.
Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$.  The distance from a point $(X,Y,Z)$ to a plane with equation $Ax+By+Cz+D=0$ is 
\[\frac{AX+BY+CZ+D}{\sqrt{A^2+B^2+C^2}},\]
so the (directed) distance from any point $(x,y,z)$ to the plane is $ax+by+cz+d$.  So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$, and by rearranging and summing, \[(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100.\]
Solving the equation is easier if we substitute $11-d=y$, to get $3y^2+2=100$, or $y=\sqrt {98/3}$.  The distance from the origin to the plane is simply $d$, which is equal to $11-\sqrt{98/3} =(33-\sqrt{294})/3$, so $33+294+3=\boxed{330}$.