imc

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 $△ABD.$ The same concept applies to the other side lengths. $AD=3$ and $BC=\sqrt{1^2+2^2}=\sqrt{5}$. Then $EF=HG=\frac{3}{2}$ and $EH=FG=\frac{\sqrt{5}}{2}$. The line segments lie on perpendicular planes so quadrilateral $EFGH$ is a rectangle. The area is $$EF\cdot FG=\frac{3}{2}\cdot \frac{\sqrt{5}}{2}=\frac{3\sqrt{5}}{4}⟹\boxed{\frac{3\sqrt{5}}{4}}.$$