Let points $A$ = $(0 ,0 ,0)$, $B$ = $(1, 0, 0)$, $C$ = $(0, 2, 0)$, and $D$ = $(0, 0, 3)$. Points $E$, $F$, $G$, and $H$ are midpoints of line segments $\overline{BD},\text{ }  \overline{AB}, \text{ } \overline {AC},$ and $\overline{DC}$ respectively. What is the area of $EFGH$?
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{3}\qquad\textbf{(C)}\ \frac{3\sqrt{5}}{4}\qquad\textbf{(D)}\ \sqrt{3}\qquad\textbf{(E)}\ \frac{2\sqrt{7}}{3}$
Consider a tetrahedron with vertices at $A,B,C,D$ on the $xyz$-plane. The length of $EF$ is just one-half of $AD$ because it is the midsegment of $\triangle ABD.$ The same concept applies to the other side lengths. $AD=3$ and $BC=\sqrt{1^2+2^2}=\sqrt{5}$. Then $EF=HG=\frac32$ and $EH=FG=\frac{\sqrt{5}}{2}$. The line segments lie on perpendicular planes so quadrilateral $EFGH$ is a rectangle. The area is
\[EF \cdot FG = \frac 32 \cdot \frac{\sqrt{5}}{2} = \frac{3\sqrt 5} 4\implies \boxed{\textbf C}.\]