jmc

Let $A,B,C$ and $E$ be the centers of the three small spheres and the large sphere, respectively. Then $△ABC$ is equilateral with side length 2. If $D$ is the intersection of the medians of $△ABC$, 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 $$\boxed{3+\frac{\sqrt{69}}{3}}.$$