6. Through the vertex $D$ of the tetrahedron $ABCD$, a sphere with radius 1 is drawn. This sphere is tangent to the circumscribed sphere of the tetrahedron $ABCD$ at point $D$, and is also tangent to the plane $ABC$. 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 $ABCD$ is ( ).
(A) 2
(B) $2\sqrt{2}$
(C) 3
(D) $2\sqrt{3}$
6. C.

Draw a perpendicular from point $D$ to plane $ABC$, with the foot of the perpendicular being $H$. Draw $DE \perp AB$, with the foot of the perpendicular being $E$, and draw $DF \perp AC$, with the foot of the perpendicular being $F$. Then $HE \perp AB, HF \perp AC$, and
$$
AE = AF = AD \cos 45^{\circ} = \sqrt{6}.
$$

Since $\triangle AEH \cong \triangle AFH$, then
$$
\begin{array}{l}
\angle HAE = 30^{\circ}, AH = \frac{AE}{\cos 30^{\circ}} = 2\sqrt{2}, \\
DH = \sqrt{AD^2 - AH^2} = 2.
\end{array}
$$

Thus, $DH$ is the diameter of a sphere with radius 1. The center $O$ of the circumscribed sphere of tetrahedron $ABCD$ lies on the extension of $DH$. We have $r^2 = (r-2)^2 + (2\sqrt{2})^2$, solving for $r$ gives $r = 3$.