3. In $\triangle A B C$, let $A D$ be the altitude, and $B E$ be the angle bisector. If $B C=6, C A=7, A B=8$, then $D E=$ $\qquad$
3. As shown in the figure, from the given information, we have $\frac{B}{A}=\frac{BC}{AB}$. Therefore, we can derive
$$
\begin{array}{l}
\frac{CE}{CE+EA}=\frac{BC}{BC+AB}. \\
\therefore CE=7 \times \frac{6}{14}=3. \\
\text{Also, } \cos C \\
=\frac{AC^2+BC-AB^2}{2 \cdot AC \cdot BC} \\
=\frac{7^2+6^2-8^2}{2 \times 7 \times 6}=\frac{1}{4}, \\
\therefore DE^2=CE^2+DC^2-2 \cdot CE \cdot DC \cdot \cos C \\
=3^2+\left(7 \times \frac{1}{4}\right)^2-2 \times 3 \times 7 \times \frac{1}{4} \times \frac{1}{4}=\frac{151}{16}.
\end{array}
$$

Therefore, $DE=\frac{\sqrt{151}}{4}$.