Problem 692

1. 12

Solution: From $\triangle B E G \backsim \triangle D A G$,
we get $D G: G B=A D: B E$
$$\begin{array}{l} \quad=2: 1, \\ \therefore \quad D B=3 G B . \end{array}$$

Connecting $D E$, then
$$\begin{aligned} S_{A B C D}= & 2 S_{\triangle B C D}=2 \times 2 S_{\triangle 1 \cdots 1} \\ = & 4 \times 3 S_{\triangle B G E}=12 . \\ & -\frac{b}{a^{\prime}}=6, \quad \frac{c}{a^{\prime}}=8 . \quad \text { Therefore, } \\ & \frac{b}{c}=-\frac{3}{4} . \end{aligned}$$

Since the sign of the linear coefficient $b$ does not change the value of the discriminant, therefore, Yi must have misread the sign of $a$ or $c$. Thus,
$$\frac{c}{a}=4 \text {. }$$

From (1) and (2), we get $\frac{b}{a}=-3$. Therefore,
$$\frac{2 b+3 c}{a}=-6+12=6 .$$