smc

The triangle is placed on the sphere so that its three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use $\text{area}=\text{inradius}\cdot \text{semiperimeter}$. The area of the triangle can be found by drawing an altitude from the vertex between sides with length $15$ to the midpoint of the side with length $24$. The Pythagorean triple $9$ - $12$ - $15$ allows us easily to determine that the base is $24$ and the height is $9$. The formula $\frac{\text{base}\cdot \text{height}}{2}$ can also be used to find the area of the triangle as $108$, while the semiperimeter is simply $\frac{15+15+24}{2}=27$. After plugging into the equation, we thus get $108=\text{inradius}\cdot 27$, so the inradius is $4$. Now, let the distance between $O$ and the triangle be $x$. Choose a point on the incircle and denote it by $A$. The distance $OA$ is $6$, because it is just the radius of the sphere. The distance from point $A$ to the center of the incircle is $4$, because it is the radius of the incircle. By using the Pythagorean Theorem, we thus find $x=\sqrt{6^2-4^2}=\sqrt{20}=\boxed{2\sqrt{5}}$.