2019 AMC 10A Problem 21

Take a cross-section perpendicular to the triangle plane

The inradius of the triangle controls the tangent distance to the sphere

The altitude to the side of length $24$ is $$\sqrt{15^2-12^2}=9,$$ so the triangle has area $\frac12\cdot24\cdot9=108$ and semiperimeter $27$. Its inradius is therefore $$r=\frac{108}{27}=4.$$

Let $P$ be the perpendicular projection of $O$ onto the triangle’s plane, and let $d=OP$. Because all three side-lines are tangent to the sphere, $P$ is the incenter and its perpendicular distance to each side is $4$. The distance from $O$ to each side-line is the sphere’s radius, $6$, so the Pythagorean theorem gives $$d^2+4^2=6^2.$$ Hence $d=2\sqrt5.$

Thus, D is the correct answer.