4. In the tetrahedron $A-BCD$, $\angle BAC = \angle CAD = \angle DAB = 60^{\circ}, AB = 1, AC = 2, AD = 3$. Then the volume of the tetrahedron is ( ).
(A) $\frac{\sqrt{2}}{2}$
(B) $\frac{\sqrt{3}}{3}$
(C) $\frac{\sqrt{6}}{3}$
(D) Other
4. A

Let $A C^{\prime}=A D^{\prime}=1$, then $A-B C^{\prime} D^{\prime}$ is a regular tetrahedron, with height $h=\frac{\sqrt{6}}{3}$. That is, the distance from point $B$ to plane $A C D$ is $h=\frac{\sqrt{6}}{3}$. Also, $S_{\triangle A C D}=\frac{3}{2} \sqrt{3}$, thus
$$
V_{A-B C D}=V_{B-A C D}=\frac{1}{3} S_{\triangle A C D} \cdot h=\frac{\sqrt{2}}{2} .
$$