Three mutually tangent spheres of radius 1 rest on a horizontal plane.  A sphere of radius 2 rests on them.  What is the distance from the plane to the top of the larger sphere?
Let $A,B,C$ and $E$ be  the centers of the three small spheres and the large sphere, respectively. Then $\triangle ABC$ is equilateral with side length 2. If $D$ is the intersection of the medians of $\triangle ABC$, then $E$ is directly above $D$. Because $AE=3$ and $AD=2\sqrt{3}/3$, it follows that \[
DE= \sqrt{3^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}}= \frac{\sqrt{69}}{3}.
\]Because $D$ is 1 unit above the plane and the top of the larger sphere is 2 units above $E$, the distance from the plane to the top of the larger sphere is \[
\boxed{3+ \frac{\sqrt{69}}{3}}.
\][asy]
pair A,B,C,D;
A=(10,0);
B=(0,0);
C=(5,8.7);
D=(5,2.9);
draw(Circle(A,5),linewidth(0.7));
draw(Circle(B,5),linewidth(0.7));
draw(Circle(C,5),linewidth(0.7));
draw(A--B--C--cycle,linewidth(0.7));
draw(C--D--B,linewidth(0.7));
draw(D--A,linewidth(0.7));
label("1",(2.5,0),S);
label("1",(7.5,0),S);
label("$A$",(10,0),SE);
label("$B$",(0,0),SW);
label("$C$",(5,8.7),N);
label("$D$",(3,4),S);
[/asy]