3. In the acute triangle $\triangle A B C$, $\angle A=60^{\circ}, A D \perp B C$ at point $D, B D=1, C D=2$. Then the area of $\triangle A B C$ is $\qquad$
3. $\frac{3(\sqrt{3}+\sqrt{11})}{4}$.

As shown in Figure 4, draw $B E \perp$
$A C$ at point $E$, intersecting $A D$ at
point $F$. Then
$$
\angle 1=\angle 2 \text {. }
$$

From $\triangle A F E \backsim$
$\triangle B C E$
$$
\begin{aligned}
\Rightarrow & \frac{A F}{B C}=\frac{A E}{B E} \\
\Rightarrow A F & =\frac{A E}{B E} \cdot B C=B C \cot \angle B A E=\sqrt{3} .
\end{aligned}
$$

From $\triangle B F D \backsim \triangle A C D$
$$
\begin{array}{l}
\Rightarrow \frac{F D}{C D}=\frac{B D}{A D} \Rightarrow \frac{A D-A F}{C D}=\frac{B D}{A D} \\
\Rightarrow \frac{A D-\sqrt{3}}{2}=\frac{1}{A D} \Rightarrow A D^{2}-\sqrt{3} A D-2=0
\end{array}
$$
$\Rightarrow A D=\frac{\sqrt{3}+\sqrt{11}}{2}$ or $A D=\frac{\sqrt{3}-\sqrt{11}}{2}$ (discard).
Thus, $S_{\triangle A B C}=\frac{1}{2} B C \cdot A D=\frac{3(\sqrt{3}+\sqrt{11})}{4}$.