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If the volume of a tetrahedron whose conterminous edges are $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$ is 24 cubic units, then the volume of parallelepiped whose coterminous edges are $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ is
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The correct answer is:
72 cubic units
As per data given, we write
$$
\begin{aligned}
& 24=\frac{1}{6}\{(\overline{\mathrm{a}}+\overline{\mathrm{b}}) \cdot[(\overline{\mathrm{b}}+\overline{\mathrm{c}}) \times(\overline{\mathrm{c}}+\overline{\mathrm{a}})]\} \\
& =\frac{1}{6}\{(\overline{\mathrm{a}}+\overline{\mathrm{b}}) \cdot[(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+(\overline{\mathrm{b}}+\overline{\mathrm{a}})+(\overline{\mathrm{c}} \times \overline{\mathrm{a}})]\} \quad \ldots[\because \overline{\mathrm{c}} \times \overline{\mathrm{c}}=0] \\
& \therefore 144=[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]+[\overline{\mathrm{b}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]+[\overline{\mathrm{b}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{a}})]+[\overline{\mathrm{a}} \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}})]+[\overline{\mathrm{b}} \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}})] \\
& =[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]+0+0+0+0[\overline{\mathrm{b}}(\overline{\mathrm{c}} \times \overline{\mathrm{a}})] \\
& =2[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})] \quad \ldots[\because \overline{\mathrm{b}}(\overline{\mathrm{c}} \times \overline{\mathrm{a}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})] \\
& \therefore \overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=72, \text { i.e. volume of parallelepiped with } \\
& \text { conterminous edges } \overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}} .
\end{aligned}
$$
$$
\begin{aligned}
& 24=\frac{1}{6}\{(\overline{\mathrm{a}}+\overline{\mathrm{b}}) \cdot[(\overline{\mathrm{b}}+\overline{\mathrm{c}}) \times(\overline{\mathrm{c}}+\overline{\mathrm{a}})]\} \\
& =\frac{1}{6}\{(\overline{\mathrm{a}}+\overline{\mathrm{b}}) \cdot[(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+(\overline{\mathrm{b}}+\overline{\mathrm{a}})+(\overline{\mathrm{c}} \times \overline{\mathrm{a}})]\} \quad \ldots[\because \overline{\mathrm{c}} \times \overline{\mathrm{c}}=0] \\
& \therefore 144=[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]+[\overline{\mathrm{b}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]+[\overline{\mathrm{b}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{a}})]+[\overline{\mathrm{a}} \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}})]+[\overline{\mathrm{b}} \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}})] \\
& =[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]+0+0+0+0[\overline{\mathrm{b}}(\overline{\mathrm{c}} \times \overline{\mathrm{a}})] \\
& =2[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})] \quad \ldots[\because \overline{\mathrm{b}}(\overline{\mathrm{c}} \times \overline{\mathrm{a}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})] \\
& \therefore \overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=72, \text { i.e. volume of parallelepiped with } \\
& \text { conterminous edges } \overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}} .
\end{aligned}
$$
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