Solution:

Note that $EF = \dfrac{1}{2}AD$ since it is a midsegment of $\triangle ABD.$ Similarly, $HG = \dfrac{1}{2}AD$ and $FG = \dfrac{1}{2}BC.$

We also have that $\overline{EF}$ and $\overline{HG}$ are perpendicular to the $xy$-plane, which means that they are perpendicular to $\overline{FG}$ and $\overline{EH}.$

This tells us that $EFGH$ is rectangle since $EF = HG.$ We have $EF = \dfrac{1}{2} \cdot 3 = \dfrac{3}{2}.$

We also have that $$FG = \dfrac{1}{2} \sqrt{1^2 + 2^2} = \dfrac{\sqrt{5}}{2}.$$

The area of $EFGH$ is then $$EF \cdot FG = \dfrac{3}{2} \cdot \dfrac{\sqrt{5}}{2} = \dfrac{3\sqrt{5}}{4}.$$

Thus, C is the correct answer.