6 In the tetrahedron $ABCD$, it is known that $\angle ADB=\angle BDC=\angle CDA=$ $60^{\circ}, AD=BD=3, CD=2$, then the radius of the circumscribed sphere of the tetrahedron $ABCD$ is $\qquad$.
6. $\sqrt{3}$ Hint: Let the center of the circumsphere of tetrahedron $ABCD$ be $O$, then $O$ lies on the perpendicular line through the circumcenter $N$ of $\triangle ABD$ and perpendicular to the plane $ABD$. Given that $\triangle ABD$ is an equilateral triangle, point $N$ is the centroid of $\triangle ABD$. Let $P$ and $M$ be the midpoints of $AB$ and $CD$ respectively, then $N$ lies on $DP$, and $ON \perp DP, OM \perp CD$.

Since $\angle CDA = \angle CDB = \angle ADB = 60^{\circ}$, let the angle between $CD$ and the plane $ABD$ be $\theta$. We can find that $\cos \theta = \frac{1}{\sqrt{3}}, \sin \theta = \frac{\sqrt{2}}{\sqrt{3}}$.

In $\triangle DMN$, $DM = \frac{1}{2} CD = 1$, $DN = \frac{2}{3} \cdot DP = \frac{2}{3} \cdot \frac{\sqrt{3}}{2} \cdot 3 = \sqrt{3}$. By the cosine rule,
$$
MN^2 = 1^2 + (\sqrt{3})^2 - 2 \cdot 1 \cdot \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 2,
$$

thus $MN = \sqrt{2}$. The diameter of the circumcircle of quadrilateral $DMON$
$$
OD = \frac{MN}{\sin \theta} = \frac{\sqrt{2}}{\frac{\sqrt{2}}{\sqrt{3}}} = \sqrt{3}.
$$

Therefore, the radius $R$ of the sphere $O$ is $\sqrt{3}$.