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(C) The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by $V_{tet} = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.Giv

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$36$

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(C) The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by $V_{tet} = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.Given $V_{tet} = 3$,we have $\frac{1}{6} |[\vec{a} \vec{b} \vec{c}]| = 3$,which implies $|[\vec{a} \vec{b} \vec{c}]| = 18$.Let $V_{par}$ be the volume of the parallelepiped formed by the vectors $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$.The volume is given by the scalar triple product: $V_{par} = |[(\vec{a} + \vec{b}) \quad (\vec{b} + \vec{c}) \quad (\vec{c} + \vec{a})]|$.Using the property of scalar triple products,we expand this as:$[\vec{a} + \vec{b} \quad \vec{b} + \vec{c} \quad \vec{c} + \vec{a}] = (\vec{a} + \vec{b}) \cdot ((\vec{b} + \vec{c}) 	imes (\vec{c} + \vec{a}))$$= (\vec{a} + \vec{b}) \cdot (\vec{b} 	imes \vec{c} + \vec{b} 	imes \vec{a} + \vec{c} 	imes \vec{c} + \vec{c} 	imes \vec{a})$$= (\vec{a} + \vec{b}) \cdot (\vec{b} 	imes \vec{c} + \vec{b} 	imes \vec{a} + \vec{c} 	imes \vec{a})$$= \vec{a} \cdot (\vec{b} 	imes \vec{c}) + \vec{a} \cdot (\vec{b} 	imes \vec{a}) + \vec{a} \cdot (\vec{c} 	imes \vec{a}) + \vec{b} \cdot (\vec{b} 	imes \vec{c}) + \vec{b} \cdot (\vec{b} 	imes \vec{a}) + \vec{b} \cdot (\vec{c} 	imes \vec{a})$Since the scalar triple product is zero if any two vectors are identical,we get:$= [\vec{a} \vec{b} \vec{c}] + 0 + 0 + 0 + 0 + [\vec{b} \vec{c} \vec{a}]$$= [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{c}] = 2[\vec{a} \vec{b} \vec{c}]$Thus,$V_{par} = 2 	imes 18 = 36$.

The volume of the tetrahedron formed by the vectors $\vec{a}, \vec{b}, \vec{c}$ is $3$. Then the volume of the parallelepiped formed by the coterminous edges $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$ is:

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