Solution:

We claim that triangles $ABC$ and $ABD$ are perpendicular to each other.

We can show this be dropping the altitudes from $C$ to $AB$ and from $D$ to $AB$ in each triangle.

Since $AC = AD$ and $BC = BD,$ we have that the feet of these altitudes will coincide at point $P.$

Then we have that $$CP = DP = \dfrac{3 \cdot 4}{5} = \dfrac{12}{5}.$$ We then have that $CD = CP\sqrt{2},$ which shows that $CPD$ is an isosceles right triangle.

This proves the above claim. Finally, the volume of the tetrahedron is $$\dfrac{1}{3}[ABD] \cdot CP = \dfrac{6}{3} \cdot \dfrac{12}{5} = \dfrac{24}{5}.$$

Thus, C is the correct answer.