How to determine volume of parallelepiped by 4 points Let points $(0,0,0), (1,2,x), (-2,1,0)$ and $(1,1,3)$ be at four corners of a parallelpiped. Determine
the volume of the parallelepiped  by using the determinant in terms of $x$. For what value of $x$ is
the volume $0$ . Interpret this case geometrically. 
I'm new in linear algebra. 
how just 4 points can define a parallelepiped ?  and how i can value of $x$ that will make volume $0$ ?
 any hint?
 A: If $u,v,w$ are three vectors in $\mathbb{R}^3$ that don't lie on the same plane, $|<u,v\times w>|=|\det(u,v,w)|$ gives you the volumn of the parallelepiped. 
A: Three points define a flat parallelogram, the fourth point will make sure how that parallelogram gets extruded into the third dimension.
For the volume you need to derive three vectors, by determing the difference vectors 
$$
u = \vec{PQ} = Q - P
$$ 
between two points each. Just use the point at 0 as second point, so it gets easy and take the absolute value of the determinant of those vectors:
$$
V(x) = \mbox{abs}\left(\left|
\begin{matrix}
1 & -2 & 1 \\
2 & 1 & 1 \\
x & 0 & 3
\end{matrix}
\right|\right)
= \mbox{abs}(-3x + 15)
$$
The root $x$ of $V(x)$ should by easy to compute, the geometric implication as well.
Hint: Add $5/3$ times the second vector to $1/3$ times the third vector. 
A: Maybe a picture helps (from Wikipedia):

If $a=(1,2,x)$, $b=(-2,1,0)$, $c=(1,1,3)$, then this is your parallelpiped.
You can convince yourself, that its volume is given by the determinant:
$$\left|\det\left(\begin{matrix}1 & -2 & 1\\2 & 1 & 1\\x & 0 & 3\end{matrix}\right)\right|$$
