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(A) Given that the volume of the parallelepiped formed by vectors $\bar{a}, \bar{b}, \bar{c}$ is $12$. Thus,the scalar triple product is $[\bar{a} \ba

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$4 	ext{ (units)}^3$

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(A) Given that the volume of the parallelepiped formed by vectors $\bar{a}, \bar{b}, \bar{c}$ is $12$. Thus,the scalar triple product is $[\bar{a} \bar{b} \bar{c}] = 12$. The volume of a tetrahedron with coterminous edges $\bar{u}, \bar{v}, \bar{w}$ is given by $\frac{1}{6} |[\bar{u} \bar{v} \bar{w}]|$. Here,the edges are $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$. We calculate the scalar triple product: $[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}] = (\bar{a}+\bar{b}) \cdot ((\bar{b}+\bar{c}) 	imes (\bar{c}+\bar{a}))$ $= (\bar{a}+\bar{b}) \cdot (\bar{b} 	imes \bar{c} + \bar{b} 	imes \bar{a} + \bar{c} 	imes \bar{c} + \bar{c} 	imes \bar{a})$ Since $\bar{c} 	imes \bar{c} = 0$,this simplifies to: $= (\bar{a}+\bar{b}) \cdot (\bar{b} 	imes \bar{c} + \bar{b} 	imes \bar{a} + \bar{c} 	imes \bar{a})$ $= \bar{a} \cdot (\bar{b} 	imes \bar{c}) + \bar{a} \cdot (\bar{b} 	imes \bar{a}) + \bar{a} \cdot (\bar{c} 	imes \bar{a}) + \bar{b} \cdot (\bar{b} 	imes \bar{c}) + \bar{b} \cdot (\bar{b} 	imes \bar{a}) + \bar{b} \cdot (\bar{c} 	imes \bar{a})$ $= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + [\bar{b} \bar{c} \bar{a}]$ $= [\bar{a} \bar{b} \bar{c}] + [\bar{a} \bar{b} \bar{c}] = 2[\bar{a} \bar{b} \bar{c}] = 2(12) = 24$. Therefore,the volume of the tetrahedron is $\frac{1}{6} 	imes 24 = 4 	ext{ (units)}^3$.

If the volume of the parallelepiped whose coterminous edges are along the vectors $\bar{a}, \bar{b}, \bar{c}$ is $12$,then the volume of the tetrahedron whose coterminous edges are $\bar{a}+\bar{b}, \bar{b}+\bar{c}$ and $\bar{c}+\bar{a}$ is

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