(12) The correct answer is $(C)$.

Since $A G$ is half of $A B$, the area of triangle $A C G$ is half of that of $A B C$ (indeed, the height from $C$ is the same for both triangles). Similarly, the area of $D E G$ is $1 / 3$ of the area of $A C G$. Therefore, the area of $D E G$ is $\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}$ of the area of triangle $A B C$.

Problem proposed by Matteo Protopapa.
12. The area of triangle $ABC$ is $60 \mathrm{~m}^{2}$. Let $D, E$ be the points on the side $AC$ such that $AD=DE=EC$ and let $F, G, H$ be the points on $AB$ such that $AF=FG=GH=HB$. What is the area of triangle $DEG$?
(A) $20 \mathrm{~m}^{2}$
(B) $15 \mathrm{~m}^{2}$
(C) $10 \mathrm{~m}^{2}$
(D) $12 \mathrm{~m}^{2}$
(E) $18 \mathrm{~m}^{2}$