12. As shown in Figure 3, $\triangle A B C$ is an equilateral triangle, point $C$ is on the side $D E$ of rectangle $A B D E$, the inradius of $\triangle A B C$ is 1. Then the diameter of the circumcircle of rectangle $A B D E$ is $\qquad$
12. $\sqrt{21}$.

Let point $O$ be the center of the incircle of $\triangle ABC$, and $P$ be the point where the incircle touches side $AB$. Connect $OB$ and $OP$. Then $OB=2$, $BP=\sqrt{3}$, $AB=2\sqrt{3}$.
Thus, $PC=3$.
Since quadrilateral $ABDE$ is a rectangle and $CP \perp AB$, we have $AE=3$.

By the Pythagorean theorem,
$$
BE^{2}=EA^{2}+AB^{2}=3^{2}+(2\sqrt{3})^{2}=21 \text{. }
$$

Therefore, the diameter of the larger circle is $\sqrt{21}$.