2. In a tetrahedron $ABCD$ with volume 12, points $E$, $F$, and $G$ are on edges $AB$, $BC$, and $AD$ respectively, such that $AE = 2EB$, $BF = FC$, and $AG = 2GD$. A plane through points $E$, $F$, and $G$ intersects the tetrahedron in a section $EFGH$, and the distance from point $C$ to this section is 1. The area of this section is $\qquad$
2. 7 .

As shown in Figure 3, it is easy to know that $G E / /$ $D B$. Since $D B / /$ plane $E F H G$, then $H F$ // BD. Therefore, $H$ is also the midpoint of $D C$.

Let the distances from points $A, B, C, D$ to the section $E F H G$ be $h_{a}, h_{b}, h_{c}, h_{d}$, respectively. We have
$$
\begin{array}{l}
h_{b}=h_{d}, h_{r}=h_{b}, \\
h_{a}=2 h_{b} .
\end{array}
$$

Since $h_{c}=1$, then $h_{b}=h_{d}=1, h_{a}=2$.
Since $S_{i n k}=\frac{1}{6} S_{\text {Mec }}$, then $V_{D-B E F}=\frac{1}{6} V_{\text {Mec }}$.

Since $S_{\triangle A F:}=\frac{1}{2} S_{\triangle A B C}$, then $V_{H-\triangle F C}=-\frac{1}{4} V_{\text {Hexa } A B C C D}=3$.

Let the area of the section $E F H G$ be $x$, then $V_{A-E F H G}=\frac{2}{3} x, V_{D-E F H G}=\frac{1}{3} x$.

As shown in Figure 3, we have $12=-\frac{2}{3},-3+2+\frac{1}{3} x$.
Therefore, $x=7$.