smc

Drop altitudes of triangle $ABC$ and triangle $ABD$ down from $C$ and $D$, respectively. Both will hit the same point; let this point be $T$. Because both triangle $ABC$ and triangle $ABD$ are 3-4-5 triangles, $CT=DT=\frac{3\cdot 4}{5}=\frac{12}{5}$. Because $C{T}^2+D{T}^2=2{\left(\frac{12}{5}\right)}^2={\left(\frac{12}{5}\sqrt{2}\right)}^2=C{D}^2$, it follows that the $CTD$ is a right triangle, meaning that $\angle CTD=90^{\circ }$, and it follows that planes $ABC$ and $ABD$ are perpendicular to each other. Now, we can treat $ABC$ as the base of the tetrahedron and $TD$ as the height. Thus, the desired volume is $$V=\frac{1}{3}bh=\frac{1}{3}\cdot [ABC]\cdot TD=\frac{1}{3}\cdot 6\cdot \frac{12}{5}=\frac{24}{5}$$ which is answer $\boxed{\frac{24}{5}}$