7. Through the vertex $D$ of the tetrahedron $ABCD$, draw a sphere with radius 1. This sphere is tangent to the circumscribed sphere of the tetrahedron $ABCD$ at point $D$, and is also tangent to the face $ABC$. If $AD=2\sqrt{3}, \angle BAD=\angle CAD=45^{\circ}, \angle BAC=60^{\circ}$, then the radius $r$ of the circumscribed sphere of the tetrahedron is $\qquad$.
7. 3 Detailed Explanation: Draw a perpendicular line from point $D$ to plane $A B C$, with the foot of the perpendicular being $H$. Draw $D E \perp A B$, with the foot of the perpendicular being $E$, and draw $D F \perp A C$, with the foot of the perpendicular being $F$. 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 $\angle H A E=30^{\circ}$, $A H=\frac{A E}{\cos 30^{\circ}}=2 \sqrt{2}$, $D H=\sqrt{A D^{2}-A H^{2}}=2$, so $D H$ is the diameter of a sphere with radius 1. Therefore, the center $O$ of the circumscribed sphere of tetrahedron $A B C D$ lies on the extension of $D H$, and we have $r^{2}=(r-2)^{2}+(2 \sqrt{2})^{2}$, solving for $r$ gives $r=3$.