11. In $\triangle A B C$, $B C=12$, the height $h_{a}=8$ on side $B C$, $h_{b}$ and $h_{c}$ are the heights on sides $C A$ and $A B$ respectively. Then the maximum value of the product $h_{b} h_{c}$ is $\qquad$.
11. $\frac{2304}{25}$.

As shown in Figure 4, let $B C=a, C A=b, A B=c$. From the area relationship, we have $b h_{b}=c h_{c}=a h_{a}=b c \sin \angle B A C$.
Thus, $h_{b} h_{c}=\frac{a^{2} h_{a}^{2}}{b c}=a h_{a} \sin \angle B A C=96 \sin \angle B A C$.
Clearly, vertex $A$ lies on a line $l$ that is 8 units away from and parallel to $B C$.

Construct the perpendicular bisector $A_{0} D_{0}$ of $B C$ intersecting $l$ at point $A_{0}$. Then,
$\angle B A C \leqslant \angle B A_{0} C$.
Since $A_{0} D_{0}>\frac{1}{2} B C$, we have $\angle B A_{0} C<90^{\circ}$.
Therefore, $\sin \angle B A C \leqslant \sin \angle B A_{0} C$.
Also, $\sin \angle B A_{0} C=\sin 2 \angle B A_{0} D_{0}=2 \times \frac{3}{5} \times \frac{4}{5}=\frac{24}{25}$, hence $h_{b} h_{c} \leqslant 96 \times \frac{24}{25}=\frac{2304}{25}$.