Let $A=(0,0,0)$, $B=(4,0,0)$, $C=(0,3,0)$, and $D=(0,0,12)$. Points $E$, $F$, $G$, $H$ are the midpoints of $\overline{BD}$, $\overline{AB}$, $\overline{AC}$, $\overline{DC}$ respectively. What is the area of quadrilateral $EFGH$?

1. $E=(2,0,6)$, $F=(2,0,0)$, $G=(0,\frac32,0)$, $H=(0,\frac32,6)$.
2. $\overrightarrow{FE}=(0,0,6)=\overrightarrow{GH}$, so $EFGH$ is a parallelogram; $\overrightarrow{FG}=(-2,\frac32,0)$ has length $\frac52$, and it is horizontal while $\overrightarrow{FE}$ is vertical. So $EFGH$ is a rectangle with sides $6$ and $\frac52$.
3. Its area is $6\cdot\frac52=\boxed{15}$.