Let ABCD be a regular tetrahedron. Suppose points X, Y and Z lie on rays AB, AC and AD respectively such that $XY=YZ=7 and XZ=5.$ The lengths of AX, AY and AZ are all distinct. The volume of tetrahedron AXYZ is

  
Let  $AX=\alpha ,AY=\beta$ and  $AZ=\gamma$
The volume of tetrahedron AX YZ =$\frac{1}{6\sqrt{2}}\alpha \beta \gamma$
$$\begin{gathered} {\alpha }^2+{\beta }^2-\alpha \beta =49 ------- \left(1\right) \\ {\beta }^2+{\gamma }^2-\beta \gamma =49 ------ \left(2\right) \\ {\alpha }^2+{\gamma }^2-\alpha \gamma =25 ------ \left(3\right) \\ 1-2\Rightarrow {\alpha }^2-{\gamma }^2-\alpha \beta +\beta \gamma =0 \\ (\alpha -\gamma )(\alpha +\gamma )-\beta (\alpha -\gamma )=0 \\ \Rightarrow (\alpha +\gamma -\beta )(\alpha -\gamma )=0\Rightarrow \alpha +\gamma =\beta \end{gathered}$$

Substitute  $\beta =\alpha +\gamma$ in (1)

$$\begin{gathered} {\alpha }^2+{(\alpha +\gamma )}^2-\alpha (\alpha +\gamma )=49 \\ \Rightarrow {\alpha }^2+{\gamma }^2+\alpha \gamma =49 ----- \left(4\right) \end{gathered}$$

From 3 and 4  $\Rightarrow \alpha \gamma =12, \alpha +\gamma =\sqrt{61}$
Volume  $=\frac{1}{6\sqrt{2}}\alpha \beta \gamma =\frac{1}{6\sqrt{2}}\left(\sqrt{61}\right)\left(12\right)=\sqrt{122}$