Of Cones and Spheres

Firstly, imagine the shapes on a 2d plane (i.e. you're looking at them from the front). The sphere will be an incircle and the cone will be an equilateral triangle of side 6 $\sqrt{3}$ . The area of the triangle = $A_{t}$ = $9\times3(\sqrt{3}$ ), semi-perimter = ${s}$ = 9 $\sqrt{3}$ . Now the radius of the incircle = ${r}$ = $\frac{A_t}{s}$ = 3 (by cancelling out common factors on the numerator and denominator) = radius of sphere

$\sin~\theta=\dfrac{3\sqrt{3}}{6\sqrt{3}}=\dfrac{1}{2}$ $\implies$ $\theta=30$

It means that the 2-dimensional view of the cone is an equilateral triangle. So $\angle DCO=30$ .

$\tan \angle DCO=\dfrac{r}{3\sqrt{3}}$ $\implies$ $r=\tan~30(3\sqrt{3})$ $\implies$ $\boxed{r=3}$

by ratio and proportion: 1/2(9-r) = r r = 3