3. Given that $\triangle A B C$ is an equilateral triangle, the ellipse $\Gamma$ has one focus at $A$, and the other focus $F$ lies on the line segment $B C$. If the ellipse $\Gamma$ passes exactly through points $B$ and $C$, then its eccentricity is
3. $\frac{\sqrt{3}}{3}$.

Let the side length of the equilateral $\triangle A B C$ be $x$, the semi-major axis of the ellipse $\Gamma$ be $a$, and the semi-focal distance be $c$.
According to the problem, we have
$$
|A C|+|C F|=|A B|+|B F|=2 a \text {. }
$$

Therefore, $F$ is the midpoint of $B C$, so
$$
x+\frac{x}{2}=2 a \Rightarrow a=\frac{3}{4} x \text {. }
$$

In the right triangle $\triangle A C F$, by the Pythagorean theorem, we get
$$
\begin{array}{l}
|A F|=\sqrt{|A C|^{2}-|C F|^{2}} \\
=\sqrt{x^{2}-\left(\frac{x}{2}\right)^{2}}=\frac{\sqrt{3}}{2} x \\
\Rightarrow c=\frac{\sqrt{3}}{4} x .
\end{array}
$$

In summary, the eccentricity of the ellipse $\Gamma$ is $\frac{c}{a}=\frac{\sqrt{3}}{3}$.