325. Tetrahedron - from progression. Twelve numbers $a_{i}$ form an arithmetic progression, so that $a_{k}+d=a_{k+1}$. Find the volume of the tetrahedron with vertices at points $\left(a_{1}^{2}, a_{2}^{2}, a_{3}^{2}\right),\left(a_{4}^{2}, a_{5}^{2}, a_{6}^{2}\right)$, $\left(a_{7}^{2}, a_{8}^{2}, a_{9}^{2}\right),\left(a_{10}^{2}, a_{11}^{2}, a_{12}^{2}\right)$.
325. The squares of any three consecutive terms of an arithmetic progression with difference $d$ satisfy the equation

$$
a_{i}^{2}-2\left(a_{i}+d\right)^{2}+\left(a_{i}+2 d\right)^{2}=2 d^{2}
$$

In our case, it takes the form

$$
x-2 y+z=2 d^{2}
$$

Therefore, all four vertices of the tetrahedron are coplanar, and its volume is zero.

We see that we could have taken $a_{i}$ which do not form an arithmetic progression together, but such that each
triple of consecutive $a_{i}$ forms an arithmetic progression with the same difference $d$.

[C. Trigg, M. M., 44, 114 (February 1971).]