11. $A D 、 B E 、 C F$ are the three medians of $\triangle A B C$. If $B C=a, C A=b, A B=c$, then $A D^{2}+B E^{2}+$ $C F^{2}=$ $\qquad$ .
11. $\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)$.

As shown in Figure 7, draw
$A P \perp B C$ at point $P$. Then we have
$$
\begin{array}{l}
A P^{2}=A C^{2}-P C^{2} \\
=A B^{2}-B P^{2}, \\
2 A P^{2}=A C^{2}+A B^{2}-P C^{2}-B P^{2} \\
=b^{2}+c^{2}-\left(\frac{a}{2}-D P\right)^{2}-\left(\frac{a}{2}+D P\right)^{2} \\
=b^{2}+c^{2}-\frac{a^{2}}{2}-2 D P^{2} .
\end{array}
$$

Thus, we have $2\left(A P^{2}+D P^{2}\right)=b^{2}+c^{2}-\frac{a^{2}}{2}$, which means
$$
A D^{2}=\frac{b^{2}}{2}+\frac{c^{2}}{2}-\frac{a^{2}}{4} \text {. }
$$

Similarly, $B E^{2}=\frac{a^{2}}{2}+\frac{c^{2}}{2}-\frac{b^{2}}{4}$,
$$
C F^{2}=\frac{a^{2}}{2}+\frac{b^{2}}{2}-\frac{c^{2}}{4} \text {. }
$$

Therefore, $A D^{2}+B E^{2}+C F^{2}=\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)$.