If the four faces of a tetrahedron are represented by the equations, $\overrightarrow{r}.(\beta \widehat{i}+\gamma \widehat{j})=0,\overrightarrow{r}.(\gamma \widehat{j}+\alpha \widehat{k})=0, \overrightarrow{r}.(\alpha \widehat{k}+\beta \widehat{i})=0$ and $\overrightarrow{r}.(\beta \widehat{i}+\gamma \widehat{j}+\alpha \widehat{k})=\lambda ,$ then volume of tetrahedron (in cubic units) (where $\beta ,\gamma ,\alpha ,\lambda$ are positive numbers) is

Cartesian equations of the planes are
  $\beta x+\gamma y=0$ ……..(i)   $\gamma y+\alpha z=0$ ……. (ii)
 $\alpha z+\beta x=0$  …….(iii) and  $\beta x+\gamma y+\alpha z=\lambda$ …….(iv)
Now, plane (iv) cuts the plane (i) at $(0,0,\frac{\lambda }{\alpha }),$ the plane (iv) cuts the plane (ii) at $(\frac{\lambda }{\beta },0,0)$ and the plane (iv) cuts the plane (iii) at $(0,\frac{\lambda }{\alpha },0).$ Thus the planes (i), (ii) and (iii) are coordinate planes intersecting at origin (0, 0, 0). Therefore the lengths of the coterminous edges of the tetrahedron are $(\frac{\lambda }{\beta },\frac{\lambda }{\gamma },\frac{\lambda }{\alpha })$