6. Spheres with radii $1,2,3$ are externally tangent to each other, and planes $\alpha$ and $\beta$ are tangent to all three spheres. Then the dihedral angle formed by plane $\alpha$ and plane $\beta$ is $\qquad$ .
$6.2 \arccos \frac{\sqrt{23}}{6}$.
According to the symmetry, the plane determined by the centers of the three spheres $\mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}$ must be the bisector plane of plane $\alpha$ and plane $\beta$.

Below, we find the area of $\triangle O_{1} O_{2} O_{3}$ and the area of the projection of $\triangle O_{1} O_{2} O_{3}$ on plane $\alpha$, $\triangle A B C$ (as shown in Figure 5). Without loss of generality, let the radii of spheres $O_{1}, O_{2}, O_{3}$ be $1, 2, 3$ respectively.
Then $O_{1} O_{2}=3$,
$$
\begin{array}{l}
O_{2} O_{3}=5, \\
O_{3} O_{1}=4 .
\end{array}
$$

Thus, $\triangle O_{1} O_{2} O_{3}$ is a right triangle.
Therefore, $S_{\triangle O_{1} O_{2} O_{3}}=6$.
In the right trapezoid $A B O_{2} O_{1}$, given $O_{1} O_{2}=3, O_{1} A=1$, $O_{2} B=2$, we have
$$
A B=\sqrt{O_{1} O_{2}^{2}-\left(O_{2} B-O_{1} A\right)^{2}}=\sqrt{8} .
$$

Similarly, $A C=\sqrt{12}, B C=\sqrt{24}$.
By the area formula of a triangle (Heron's formula), we get $S_{\triangle A B C}=\sqrt{23}$.
Let the angle between plane $\alpha$ and plane $O_{1} O_{2} O_{3}$ be $\theta$, then
$$
\cos \theta=\frac{S_{\triangle A B C}}{S_{\triangle O_{1} O_{2} O_{3}}}=\frac{\sqrt{23}}{6} \text {. }
$$

Therefore, the dihedral angle between plane $\alpha$ and plane $\beta$ is $2 \arccos \frac{\sqrt{23}}{6}$.