Given the system of equations $$\begin{cases} x-2y-z=1 \\\\x+z=-1 \\\\ -x-y+z=-1 \\\\ x-y-3z=k \end{cases} $$ where $k\in \mathbb{R}$ it is known that for $k=9$ the four planes given by the four equations above bound a tetrahedron. What is the volume of the tetrahedron?
It is not difficult to find the volume of the tetrahedron. Just use $V=\frac{1}{6}\det[\textbf{a b c}]$ where a,b,c are the vectors spanning the tetrahedron.
I am confused by one thing in this question. And that is that, equation $2$ is not a plane, it is a line so I can't see how three planes and one line will ever bound a tetrahedron. Furthermore, when I plot the equations I don't get a tetrahedron at all, I get this: -
Is there a mistake in the problem description, or am I not understanding something correctly here?