Corners of a tetrahedron

The box spanned by the $3$ edges from the origin has a determinant for its volume, and the tetrahedron fills a fixed fraction of that box.

1. The edges from the origin are $(2,1,0)$, $(1,3,1)$ and $(0,1,4)$; put them as the rows of $M=\left[\begin{array}{ccc}2& 1& 0\\ 1& 3& 1\\ 0& 1& 4\end{array}\right]$.
2. Expanding along row $1$ gives $\det (M)=2\times (3\times 4-1\times 1)-1\times (1\times 4-1\times 0)+0=22-4=18$.
3. The parallelepiped on these edges has volume $18$, and the tetrahedron on the same edges takes $\frac{1}{6}$ of it.
4. So the volume is $\frac{18}{6}=3$.

Answer: $3$