Firstly, we must choose the three edges $s$ such that they are not on the same side as to form an equilateral. For a tetrahedron $ABCD$, we can have $AB=BC=CD=s$. So we have an isosceles right angle triangle for $ABC$ and $BCD$. Now, we want to know what the right angle is for the other sides. angle ADC cannot be a right angle because then angleDAB is right as well and this makes the shape of $ABCD$ a square, which is obviously a contradiction. In addition, $AC=\sqrt{2}s > s$, so angle DCA=90 degrees. It is easy to see that angleDBA=90 as well. So taking $AB$ as the height and the base area as triangle $BCD$, we have $volume=\dfrac{s^3}{6}$

Note that there exists an isosceles right triangle, otherwise one length $s$ nukes other $4$ possible locations for other $s$, thus we have one place for two lengths $s$. Now wlog that isosceles right triangle be $BAC$ with $\angle BAC=90^\circ$ with $BA=AC=s$. Also, let the other vertices be $D,E,F$ of tetrahedrons surface, see figure below. Note that in the diagram below, we must have $BD=BF$, $CE=CF$ and $DA=AE$. Note that $DA=AE\neq s$, otherwise $BCF$ is equilateral. Hence, wlog $DB=BF=s$, thus $DA=BC=AE=\sqrt{2}s$ and $CE=CF=\sqrt{3}s$. This works and is unique. Now note that its base is $\triangle ABC$ and its height is $DB$, hence $$V=\frac{S_{ABC}\cdot BD}{3}=\frac{s^3}{6}.$$ [asy][asy] import olympiad; size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair A,B,C,D,E,F; A=(0,0);B=(0,1);C=(1,0);D=(-1,1);E=intersectionpoints(circle(A,1.41421356237),circle(C,1.73205080757))[1];F=intersectionpoints(circle(B,1),circle(C,1.73205080757))[0]; draw(A--C,red);draw(A--B,red);draw(B--D,red);draw(B--F,red); draw(B--C,blue);draw(D--A,blue);draw(A--E,blue); draw(C--E,green);draw(F--C,green); draw(rightanglemark(E,A,C,3),med);draw(rightanglemark(D,B,A,3),med);draw(rightanglemark(C,B,F,3),med);draw(rightanglemark(B,A,C,3),med); dot("$A$",A,W); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); [/asy][/asy]