Let's assume a plane
[tex]ax+by+cz=1[/tex]
with each of a, b, and c greater than 0. Then the plane has intercepts with coordinates
[tex]y=z=0 \implies ax = 1 \implies x = \dfrac1a[/tex]
[tex]x=z=0 \implies by = 1 \implies y = \dfrac1b[/tex]
[tex]x=y=0 \implies cz = 1 \implies z = \dfrac1c[/tex]
The volume of the tetrahedron is then
[tex]\displaystyle \int_0^{\frac1a} \int_0^{\frac{1-ax}b} \int_0^{\frac{1-ax-by}c} dz \, dy \, dx = \boxed{\frac1{6abc}}[/tex]
i.e. 1/6 times the product of the intercepts' non-zero coordinates
