I am given the points $A(4|3|1)$, $B(1|7|1)$, $C(-3|2|0)$, $D(0|0|0)$ and $S(0|3|4)$. I tried splitting this geometric construct up in two different tetrahedrons:
Using $\overrightarrow{DB}$ as a diagonal of the quadrilateral , I get that the the volume of the entire geometric construction is $29$. Using $\overrightarrow{AC}$ as diagonal that splits up the construction into two different tetrahedrons, I get $28$ however.
Why is that the case and what is the actual volume?
$$V=V_{DABS}+V_{DSBC}=\frac{1}{6}\left|\left(\begin{array}{cc}0 & 3 & 4\\4 & 3 & 1\\ 1 & 7 & 1 \end{array}\right)\right|+\frac{1}{6}\left|\left(\begin{array}{cc}0 & 3 & 4\\1 & 7 & 1\\ -3 & 2 & 0 \end{array}\right)\right|=29.$$
If by $V=V_{SABC}+V_{SDBC}$ we'll get another answer, it says that points $A$, $B$, $C$ and $D$ are not placed in the same plane. Now, just check it!
We have: $$V_{SABC}+V_{SDBC}+V_{ABCD}=V_{DABS}+V_{DSBC},$$ which gives $$V_{SABC}+V_{SDBC}\neq V_{DABS}+V_{DSBC}$$because $$V_{ABCD}=1.$$