3. To completely fit 4 steel balls with a radius of 1 into a container shaped as a regular tetrahedron, the minimum height of this regular tetrahedron is
A. $\frac{\sqrt{3}+2 \sqrt{6}}{3}$
B. $2+\frac{2 \sqrt{6}}{3}$
C. $4+\frac{2 \sqrt{6}}{3}$
D. $\frac{4 \sqrt{3}+2 \sqrt{6}}{3}$
3. B When four balls are pairwise tangent, and each set of three balls is simultaneously tangent to one face of a regular tetrahedron, the height of the tetrahedron is minimized. The centers of the four balls, \(O_{1}, O_{2}, O_{3}, O_{4}\), form the four vertices of a regular tetrahedron, with the edge length of the tetrahedron \(O_{1} O_{2} O_{3} O_{4}\) being 2, and its height is \(\sqrt{\left(\frac{\sqrt{3}}{2} \times 2\right)^{2}-\left(\frac{\sqrt{3}}{6} \times 2\right)^{2}}=\frac{2 \sqrt{6}}{3}\),
thus the height of the tetrahedron in the problem is \(2+\frac{2 \sqrt{6}}{3}\).