Volume of tetrahedron detected by four planes in $\mathbb R^3$

Question

Given the equation of $4$ planes in $\mathbb R^3$, find the volume of the tetrahedron. The planes are: $$\begin{align}&P1: x+3y+z=2\\&P2: x-2y+z=2\\&P3: -x+z=4\\&P4: x=3\end{align}$$ I dont even know where to start. I thought there was going to be a formula for this but idk. Maybe find some intersection points and then the vectors between those points and use the regular formula for the volume of a tetrahedron? Hope this is easier than it looks

Answer 1

The four points are $(-1,0,3)(3,0,-1)(3,-8/3,7)(3,4,7)$.

We have 3 position vectors $\bar{a}=(4,0,-4),\bar{b}=(4,-8/3,4),\bar{c}=(4,4,4)$ wrt $(-1,0,3)$.

The volume of the tetrahedron with relative position vectors $\bar{a},\bar{b},\bar{c}$ is $V=\left|{[\bar{a},\bar{b},\bar{c}]\over 6}\right|={320\over9}$.

MSE on tetrahedron volume with vectors