2004 Fermat Problem 25

A steel cube has edges of length 3 cm, and a cone has a diameter of 8 cm and a height of 24 cm. The cube is placed in the cone so that one of its interior diagonals coincides with the axis of the cone. What is the distance, in cm, between the vertex of the cone and the closest vertex of the cube?

\textbf{(A)}\ 6\sqrt{6}-\sqrt{3}\quad \textbf{(B)}\ \frac{12\sqrt{6}-3\sqrt{3}}{4}\quad \textbf{(C)}\ 6\sqrt{6}-2\sqrt{3}\quad \textbf{(D)}\ 5\sqrt{3}\quad \textbf{(E)}\ 6\sqrt{6}

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