A steel cube has edges of length $3 \mathrm{~cm}$, and a cone has a diameter of $8 \mathrm{~cm}$ and a height of $24 \mathrm{~cm}$. 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 $\mathrm{cm}$, between the vertex of the cone and the closest vertex of the cube?
(A) $6 \sqrt{6}-\sqrt{3}$
(B) $\frac{12 \sqrt{6}-3 \sqrt{3}}{4}$
(C) $6 \sqrt{6}-2 \sqrt{3}$
(D) $5 \sqrt{3}$
(E) $6 \sqrt{6}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-119.jpg?height=371&width=260&top_left_y=1517&top_left_x=1472)