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(C) The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $[\vec{a}, \vec{b}, \vec{c

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(C) The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $[\vec{a}, \vec{b}, \vec{c}] = V$. The volume of the new parallelepiped with edges $\vec{a}, \vec{b}+\vec{c}, \vec{a}+2\vec{b}+3\vec{c}$ is given by the scalar triple product $[\vec{a}, \vec{b}+\vec{c}, \vec{a}+2\vec{b}+3\vec{c}]$. Using the properties of the scalar triple product,we can express this as the determinant of the coefficients of the vectors $\vec{a}, \vec{b}, \vec{c}$: $[\vec{a}, \vec{b}+\vec{c}, \vec{a}+2\vec{b}+3\vec{c}] = \begin{vmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 2 & 3 \end{vmatrix} [\vec{a}, \vec{b}, \vec{c}]$. Calculating the determinant: $1(1 	imes 3 - 1 	imes 2) - 0 + 0 = 1(3 - 2) = 1$. Thus,the volume is $1 	imes V = V$.

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

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