Let the vectors $\vec{a}, \vec{b}, \vec{c}$ represent three coterminous edges of a parallelopiped of volume V. Then the volume of the parallelopiped, whose coterminous edges are represented by $\vec{a}, \vec{b}+\vec{c}$ and $\vec{a}+2 \vec{b}+3 \vec{c}$ is equal to :

Given that the volume $V$ of the parallelepiped formed by the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is represented by the scalar triple product $[\vec{a},\vec{b},\vec{c}]$, which is the determinant of the 3 x 3 matrix with vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ as its rows (or columns). <br/><br/>When the vectors representing the coterminous edges of the parallelepiped are $\vec{a}$, $\vec{b} + \vec{c}$, and $\vec{a} + 2\vec{b} + 3\vec{c}$, the volume $V$ of the parallelepiped is represented by : <br/><br/>$\begin{aligned} & V=[\vec{a}, \vec{b}+\vec{c}, \vec{a}+2 \vec{b}+3 \vec{c}] \\\\ & =\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 3\end{array}\right|\left[\begin{array}{ll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] \\\\ & =1(3-2)[\vec{a}, \vec{b}, \vec{c}] \\\\ & =\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=V \end{aligned}$