3. Given a tetrahedron $S-ABC$, point $A_{1}$ is the centroid of $\triangle SBC$, and $G$ is on segment $AA_{1}$, satisfying $\frac{|AG|}{|GA_{1}|}=3$. Connecting $SG$ intersects the plane of $\triangle ABC$ at point $M$, then $\frac{|A_{1}M|}{|AS|}=$ $\qquad$ .
Extend $S A_{1}$ to intersect $B C$ at point $D$, then $D$ is the midpoint of $B C$, and the plane $S A A_{1} \cap$ plane $A B C=A D$, thus $A, M, D$ are collinear.
$$
\begin{array}{l}
\overrightarrow{S A_{1}}=\frac{2}{3} \overrightarrow{S D} \Rightarrow \overrightarrow{A A_{1}}=\frac{2}{3} \overrightarrow{A D}+\frac{1}{3} \overrightarrow{A S} \\
\Rightarrow \overrightarrow{A G}=\frac{3}{4} \overrightarrow{A A_{1}}=\frac{1}{2} \overrightarrow{A D}+\frac{1}{4} \overrightarrow{A S},
\end{array}
$$

Then $\frac{1}{2} \overrightarrow{A D}=\frac{3}{4} \overrightarrow{A M} \Rightarrow \overrightarrow{A M}=\frac{2}{3} \overrightarrow{A D}$, so $M$ is the centroid of $\triangle A B C$, hence $A_{1} M / / S A \Rightarrow \frac{\left|A_{1} M\right|}{|A S|}=\frac{1}{3}$.