
The volume of the parallelepiped is the base area times the height. The base area, as you know, is the magnitude of $\vec a\times\vec b$, that is $\|\vec a\times\vec b\|$.
You also know that $\vec a\times\vec b$ is a vector perpendicular to the plane containing $\vec a,\vec b$. Thus, the height of the parallelepiped is the absolute value of the projection of $\vec c$ along $\vec a\times\vec b$, that is $\Big|\vec c\cdot\frac{\vec a\times\vec b}{\|\vec a\times\vec b\|}\Big|$ (for the derivation of this quantity, see this answer).
Multiply them together to get$$V=\Big|\vec c\cdot\frac{\vec a\times\vec b}{\|\vec a\times\vec b\|}\Big|\|\vec a\times\vec b\|=|\vec c\cdot\vec a\times\vec b|$$
$\vec c\cdot\vec a\times\vec b$ is called the scalar triple product of vectors $\vec a,\vec b,\vec c$ and denoted as $[\vec a\ \vec b\ \vec c]$. The volume of the parallelepiped formed by three vectors is the geometrical interpretation of the absolute value of their scalar triple product.
In your case, note that $\vec a,\vec b,\vec c$ are coplanar, as $\vec b-\vec c=\vec a$. The volume of the parallelepiped formed by three vectors lying in the same plane is zero. You can verify the same using scalar triple product.
(can someone tell me where I can format?)
– Zealotory Jan 11 '19 at 06:42