3. Given a regular tetrahedron $ABCD$, point $G$ is the centroid of the base $\triangle ABC$, and point $M$ lies on the line segment $DG$ such that the sine of the obtuse angle $AMB$ is $\frac{3 \sqrt{51}}{22}$. Then $\frac{DM}{MG}=$ $\qquad$.
3. 2 Detailed Explanation: From the problem, we can set $A M=B M=x, A B=1$. By the cosine theorem, we get $x=\frac{\sqrt{33}}{9}$, and $A G=G B=\frac{\sqrt{3}}{3}$. Then, $G M=\sqrt{A M^{2}-A G^{2}}=\frac{\sqrt{6}}{9}$. And $D G=\frac{\sqrt{6}}{3}$. Therefore, $\frac{D M}{M G}=\frac{\frac{\sqrt{6}}{3}-\frac{\sqrt{6}}{9}}{\frac{\sqrt{6}}{9}}=2$.