Example 3 Put a triangular wire frame with side lengths of $13, 14, 15$ on a sphere $O$ with a radius of 10, find the distance between the plane where this triangle lies and the center of the sphere $O$.
Let the incenter of the wire frame $\triangle ABC$ be $I$, and the inradius be $r$. Then, by the property of the spherical section, we know that $OI \perp$ plane $ABC$. Therefore, the distance we are looking for is the length of $OI$, and $OI=\sqrt{R^{2}-r^{2}}=\sqrt{10^{2}-r^{2}}$.
From $S_{\text {S } ABC}=\frac{1}{2}(a+b+c) r=\frac{1}{4} \sqrt{(a+b+c)(a+b-c)(c+a-b)(b+c-a)}$
we get $\quad \frac{1}{2} \cdot(13+14+15) \cdot r=\frac{1}{4} \sqrt{42 \times 14 \times 16 \times 12}$
So $r=4$
Thus, $OI=\sqrt{100-16}=2 \sqrt{21}$, hence the distance we are looking for is $2 \sqrt{21}$.