12. In the tetrahedron $ABCD$, $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$; if a sphere passing through $D$ is tangent to the plane $ABC$ and internally tangent to the circumscribed sphere of the tetrahedron with a radius of 1, then the radius of the circumscribed sphere of the tetrahedron $ABCD$ is $\qquad$.
(12) 3 Hint: Draw a perpendicular from point $D$ to plane $A B C$, with the foot of the perpendicular being $H$. Draw $D E \perp A B$, $D F \perp A C$, with the feet of the perpendiculars being $E$ and $F$, respectively. Then $H E \perp A B, H F \perp A C$ and
$$
A E=A F=A D \cos 45^{\circ}=\sqrt{6},
$$

Since $\triangle A E H \cong \triangle A F H$, then $\angle H A E=30^{\circ}$.
$$
A H=\frac{A E}{\cos 30^{\circ}}=2 \sqrt{2}, \quad D H=\sqrt{A D^{2}-A H^{2}}=2,
$$

Therefore, $D H$ is the diameter of a sphere with radius 1, so the center $O$ of the circumscribed sphere of tetrahedron $A B C D$ lies on the extension of $D H$. Let the radius of the circumscribed sphere be $r, r^{2}=(r-2)^{2}+$ $(2 \sqrt{2})^{2}$, solving this gives $r=3$.