6. In a tetrahedron $ABCD$ with volume 12, points $E, F, G$ are on edges $AB, BC, AD$ respectively, and $AE = 2EB, BF = FC, AG = 2GD$. A section $EFHG$ is made through points $E, F, G$. The distance from point $C$ to this section is 1. Then the area of this section is $\qquad$ .
6. 7 As shown in Figure (1), it is easy to know that $G E / / D B$, so $D B / /$ plane $F E H G$, then $H F / / B D$, thus, $H$ is 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, with $h_{B}=h_{C}=h_{D}=\frac{1}{2} h_{A}$. Given $h_{C}=1$, so $h_{B}=h_{D}=1, h_{A}=2$. Since $S_{\triangle B E F}=\frac{1}{6} S_{\triangle A B C}$, then $V_{D-B E F}=\frac{1}{6} V_{\text {tetrahedron } A B C D}=2$. Since $S_{\triangle A F C}=\frac{1}{2} S_{\triangle A B C}$, then $V_{H-A F C}=\frac{1}{4} V_{\text {tetrahedron } A B C D}=3$. As shown in Figure (2), let the area of the section $E F H G$ be $x$,
then $V_{A-E F H G}=\frac{2}{3} x, V_{D-E F H G}=\frac{1}{3} x$.
Since $V_{\text {tetrahedron } A B C D}=V_{A-E F H G}+V_{H-A F C}+V_{D-B E F}+V_{D-E F H G}$, i.e., $12=\frac{2}{3} x+3+2+\frac{1}{3} x$,
so $x=7$.