3. Given that the perimeter of $\triangle A B C$ is 20, the radius of the inscribed circle is $\sqrt{3}$, and $B C=7$. Then the value of $\tan A$ is
3. $\sqrt{3}$.

Let $A B=x, A C=y$.
From the given information, we have
$$
x+y=13, S_{\triangle A B C}=10 \sqrt{3} \text {. }
$$

Using the triangle area formula and the cosine rule, we get
$$
\frac{1}{2} x y \sin A=10 \sqrt{3} \text {, }
$$
$$
x^{2}+y^{2}-2 x y \cos A=49 \text {. }
$$

From $x+y=13$, we obtain
$$
x^{2}+y^{2}=169-2 x y \text {. }
$$

Substituting into equation (2), we get
$$
x y(1+\cos A)=60 \text {. }
$$
Dividing (1) by (3) gives
$$
\begin{array}{l}
\frac{\sin A}{1+\cos A}=\frac{\sqrt{3}}{3} \Rightarrow \tan \frac{A}{2}=\frac{\sqrt{3}}{3} \\
\Rightarrow \tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^{2} \frac{A}{2}}=\sqrt{3} .
\end{array}
$$