12. $S-A B C D$ is a regular quadrilateral pyramid, $P, Q$ are the midpoints of $S B, S D$ respectively, the plane through points $A, P, Q$ divides $S-$ $A B C D$ into two parts with volumes $V_{1} \cdot V_{2}\left(V_{1}<V_{2}\right)$, then the ratio $\frac{V_{2}}{V_{1}}$ equals $\qquad$ .
12. 5 As shown in the figure, let the plane $A P Q$ intersect $S C$ at $R$, and $A R$ intersect $P Q$ at $O$. By symmetry, $O$ is the midpoint of $P Q$,
thus $S O \perp$ plane $A B C D$, extending $S O$ to intersect the base $A B C D$ at $O_{1}$,
line $A O C$ intersects $\triangle S O_{1} R$, applying Menelaus' theorem we have:
$$
\frac{S R}{R C} \cdot \frac{C A}{A O_{1}} \cdot \frac{O_{1} O}{O S}=1,
$$

thus $C R=2 S R$.
Also, $V_{S-A Q R P}=V_{A-S P R}+V_{A-S Q R}$,
and $V_{A-S P R}=\frac{1}{6} V_{A-S B C}=\frac{1}{12} V_{S-A B C D}$,
similarly $V_{A-S P R}=\frac{1}{12} V_{S-A B C D}$.
Therefore, $V_{S-A P R Q}=\frac{1}{6} V_{S-A B C D}$, so $\frac{V_{2}}{V_{1}}=\frac{6-1}{1}=5$.