2004 AMC 12A Problem 22

Let the centers of the three unit spheres be $A,$ $B,$ $C,$ forming an equilateral triangle of side $2$ at height $1$ above the plane, and let $E$ be the center of the large sphere directly above the centroid $D$ of $\triangle ABC.$ The distance from a vertex to the centroid is $AD = \tfrac{2\sqrt3}{3},$ and $AE = 1 + 2 = 3,$ so $$\begin{aligned} DE &= \sqrt{3^2 - \left(\tfrac{2\sqrt3}{3}\right)^2} \\ &= \sqrt{9 - \tfrac{4}{3}} \\ &= \dfrac{\sqrt{69}}{3}. \end{aligned}$$ Since $D$ is $1$ unit above the plane and the top of the large sphere is $2$ units above $E,$ the total height is $$1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.$$ Thus, the correct answer is B.