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(C) The volume of a tetrahedron with edges $\overline{u}, \overline{v}, \overline{w}$ is given by $V = \frac{1}{6} |[\overline{u} \overline{v} \overli

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$72$ cubic units

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(C) The volume of a tetrahedron with edges $\overline{u}, \overline{v}, \overline{w}$ is given by $V = \frac{1}{6} |[\overline{u} \overline{v} \overline{w}]|$.Given edges are $\overline{a}+\overline{b}, \overline{b}+\overline{c}, \overline{c}+\overline{a}$.So,$24 = \frac{1}{6} |(\overline{a}+\overline{b}) \cdot ((\overline{b}+\overline{c}) 	imes (\overline{c}+\overline{a}))|$.$144 = |(\overline{a}+\overline{b}) \cdot (\overline{b} 	imes \overline{c} + \overline{b} 	imes \overline{a} + \overline{c} 	imes \overline{c} + \overline{c} 	imes \overline{a})|$.Since $\overline{c} 	imes \overline{c} = 0$,we have $144 = |(\overline{a}+\overline{b}) \cdot (\overline{b} 	imes \overline{c} + \overline{b} 	imes \overline{a} + \overline{c} 	imes \overline{a})|$.Expanding the scalar triple product: $144 = |[\overline{a} \overline{b} \overline{c}] + [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{c} \overline{a}] + [\overline{b} \overline{b} \overline{c}] + [\overline{b} \overline{b} \overline{a}] + [\overline{b} \overline{c} \overline{a}]|$.Terms like $[\overline{a} \overline{b} \overline{a}]$ are $0$ because two vectors are identical.Thus,$144 = |[\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{c} \overline{a}]|$.Since $[\overline{a} \overline{b} \overline{c}] = [\overline{b} \overline{c} \overline{a}]$,we get $144 = 2 |[\overline{a} \overline{b} \overline{c}]|$.Therefore,$|[\overline{a} \overline{b} \overline{c}]| = 72$,which is the volume of the parallelepiped.

If the volume of a tetrahedron whose conterminous edges are $\overline{a}+\overline{b}, \overline{b}+\overline{c}, \overline{c}+\overline{a}$ is $24$ cubic units,then the volume of the parallelepiped whose coterminous edges are $\overline{a}, \overline{b}, \overline{c}$ is

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