3. In $\triangle A B C$, the lengths of the three sides are $a=3$, $b=4, c=6, h_{a}$ represents the altitude on side $B C$, and $h_{b} 、 h_{c}$ are defined similarly. Then
$$
\left(h_{a}+h_{b}+h_{c}\right)\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right)=
$$
$\qquad$
$3.9 \frac{3}{4}$.
Let the area of $\triangle ABC$ be $S$, then we have
$$
\begin{array}{l}
S=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c} \text {. } \\
\text { Therefore, }\left(h_{a}+h_{b}+h_{c}\right)\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right) \\
=\left(\frac{2 S}{a}+\frac{2 S}{b}+\frac{2 S}{c}\right)\left(\frac{a}{2 S}+\frac{b}{2 S}+\frac{c}{2 S}\right) \\
=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(a+b+c) \\
=\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\right)(3+4+6) \\
=9 \frac{3}{4} \text {. } \\
\end{array}
$$