3. Let the orthocenter and circumcenter of $\triangle A B C$ be $H, O$, respectively, and the circumradius be 3,
$$
|O H|=1,|B C|=a,|C A|=b,|A B|=c
$$

Then $a^{2}+b^{2}+c^{2}=$ $\qquad$ .
Solution:
As shown in the figure, draw the diameter $B D$, then $D A \perp A B, C D \perp B C$, and quadrilateral $A H C D$ is a parallelogram.
$$
\text { Hence } \overrightarrow{O H}=\overrightarrow{O A}+\overrightarrow{A H}=\overrightarrow{O A}+\overrightarrow{D C}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}
$$

Also,
$$
\begin{array}{l} 
d^{2}=|\overrightarrow{O H}|^{2}=(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C})^{2}=3 R^{2}+2 R^{2}(\cos \angle B O C+\cos \angle C O A+\cos \angle A O B) \\
=3 R^{2}+2 R^{2}-a^{2}+2 R^{2}-b^{2}+2 R^{2}-c^{2}=9 R^{2}-\left(a^{2}+b^{2}+c^{2}\right)
\end{array}
$$

Thus, $a^{2}+b^{2}+c^{2}=9 R^{2}-d^{2}=80$.