Example 6 As shown in Figure $6, \odot O$ is the circumcircle of $\triangle A B C$, and point $H$ is the orthocenter of $\triangle A B C$. Let $B C=a, C A=b$, $A B=c$, and the radius of $\odot O$ be $R$. Try to express the length of $O H$ using $a, b, c, R$.

Express the length of $O H$ using $a, b, c, R$.
As shown in Figure 6, take the centroid $G$ of $\triangle ABC$, then point $G$ must lie on $OH$ (by the definition of the Euler line), and
$$
O H=3 O G \text{. }
$$

The ray $AG$ intersects $\odot O$ at point $A'$ and intersects side $BC$ at point $D$, where $D$ is the midpoint of $BC$. The line $OH$ intersects $\odot O$ at points $M$ and $N$. It is easy to see that
$$
\begin{array}{l}
D A'=\frac{B D \cdot D C}{A D}=\frac{a^{2}}{4 A D}, \\
G A'=G D+D A'=\frac{1}{3} A D+\frac{a^{2}}{4 A D}, \\
A G=\frac{2}{3} A D .
\end{array}
$$

Noting that $A D^{2}=\frac{1}{2}\left(b^{2}+c^{2}-\frac{1}{2} a^{2}\right)$, we have
$$
\begin{array}{l}
R^{2}-O G^{2}=M G \cdot G N=A G \cdot G A' \\
=\frac{2}{3} A D\left(\frac{1}{3} A D+\frac{a^{2}}{4 A D}\right) \\
=\frac{2}{9} A D^{2}+\frac{1}{6} a^{2} \\
=\frac{2}{9} \times \frac{1}{2}\left(b^{2}+c^{2}-\frac{1}{2} a^{2}\right)+\frac{1}{6} a^{2} \\
=\frac{1}{9}\left(a^{2}+b^{2}+c^{2}\right) .
\end{array}
$$

Thus, $O G^{2}=R^{2}-\frac{1}{9}\left(a^{2}+b^{2}+c^{2}\right)$,
$$
O H^{2}=9 O G^{2}=9 R^{2}-\left(a^{2}+b^{2}+c^{2}\right) \text{. }
$$

Therefore, $O H=\sqrt{9 R^{2}-\left(a^{2}+b^{2}+c^{2}\right)}$.