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If the volume of the parallelopiped whose conterminus edges are along the vectors $\bar{a}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ is 12, then the volume of the tetrahedron whose conterminus edges are $\bar{a}+\overline{\mathrm{b}}, \overline{\mathrm{b}}+\overline{\mathrm{c}}$ and $\overline{\mathrm{c}}+\overline{\mathrm{a}}$ is
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4 (units) $^{3}$
(C)
Volume of parallelepiped $=[\bar{a} \bar{b} \bar{c}]=12$
$\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=12$$\ldots(1)$
volume of tetrahedrom
$=\frac{1}{6}\left[\begin{array}{lll}
\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}
\end{array}\right]$
$=\frac{1}{6}(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]=\frac{1}{6}(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]$
$=\frac{1}{6}\{\bar{a} \cdot(\bar{b} \times \bar{c})+\bar{a}(\bar{b} \times \bar{a})+\bar{a}(\bar{c} \times \bar{a})+\bar{b} \cdot(\bar{b} \times \bar{c})+\bar{b} \cdot(\bar{b} \times \bar{a})+\bar{b}(\bar{c} \times \bar{a})\} \quad \ldots[\because \bar{c} \times \bar{c}=0]$
$=\frac{1}{6}[\bar{a} \cdot(\bar{b} \times \bar{c})+0+\bar{b} \cdot(\bar{c} \times \bar{a})]=\frac{1}{6}\{[\bar{a} \bar{b} \bar{c}]+[\bar{a} \bar{b} \bar{c}]\}$
$=\frac{2}{6}[\bar{a} \bar{b} \bar{c}]=\frac{1}{3}(12)=4$
Volume of parallelepiped $=[\bar{a} \bar{b} \bar{c}]=12$
$\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=12$$\ldots(1)$
volume of tetrahedrom
$=\frac{1}{6}\left[\begin{array}{lll}
\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}
\end{array}\right]$
$=\frac{1}{6}(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]=\frac{1}{6}(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]$
$=\frac{1}{6}\{\bar{a} \cdot(\bar{b} \times \bar{c})+\bar{a}(\bar{b} \times \bar{a})+\bar{a}(\bar{c} \times \bar{a})+\bar{b} \cdot(\bar{b} \times \bar{c})+\bar{b} \cdot(\bar{b} \times \bar{a})+\bar{b}(\bar{c} \times \bar{a})\} \quad \ldots[\because \bar{c} \times \bar{c}=0]$
$=\frac{1}{6}[\bar{a} \cdot(\bar{b} \times \bar{c})+0+\bar{b} \cdot(\bar{c} \times \bar{a})]=\frac{1}{6}\{[\bar{a} \bar{b} \bar{c}]+[\bar{a} \bar{b} \bar{c}]\}$
$=\frac{2}{6}[\bar{a} \bar{b} \bar{c}]=\frac{1}{3}(12)=4$
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