44 There exists a tetrahedron that satisfies the problem. Construct it as follows: $T_{1}$ is a regular tetrahedron with edge length $\frac{2}{3} \sqrt[4]{27}$, and the area of each of its faces is $\frac{\sqrt{3}}{4}\left(\frac{2}{3} \sqrt[4]{27}\right)^{2}=1$.
Take a regular triangle $A B C$ with an area greater than 3, and draw a perpendicular line through its center $O$. Take any point $P$ on this perpendicular line, then
$$S_{\triangle P A B}=S_{\triangle P B C}=S_{\triangle P C A}>S_{\triangle O B C}>1 .$$

However, when $P$ is sufficiently close to $O$, the volume of this tetrahedron $T_{2}$ can be arbitrarily close to zero.