3. In $\triangle A B C$, it is known that $\angle A=60^{\circ}, \angle C=$ $75^{\circ}, A B=10$, points $D, E, F$ are on sides $A B, B C, C A$ respectively. Then the minimum perimeter of $\triangle D E F$ is $\qquad$
$3.5 \sqrt{6}$.
As shown in Figure 5, construct the symmetric points $P$ and $Q$ of point $E$ with respect to $AB$ and $AC$, and connect $AE$, $AP$, $AQ$, $DP$, $FQ$, and $PQ$.

Then $\angle PAQ = 120^{\circ}$, and $AP = AQ = AE$.
Construct $AH \perp BC$ at point $H$.
Then the perimeter of $\triangle DEF$
$$
\begin{array}{l}
l = DE + DF + EF = PD + DF + FQ \\
\geqslant PQ = \sqrt{3} AP = \sqrt{3} AE \geqslant \sqrt{3} AH .
\end{array}
$$

Since $AH = 10 \cos B = 10 \cos 45^{\circ} = 5 \sqrt{2}$, we have $l \geqslant \sqrt{3} \times 5 \sqrt{2} = 5 \sqrt{6}$, with equality holding if and only if point $E$ coincides with $H$, and points $P$, $D$, $F$, and $Q$ are collinear.