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If $[\bar{a} \bar{b} \bar{c}]=3$, then the volume of the parallelopiped with $2 \bar{a}+\bar{b}, 2 \bar{b}+\bar{c}, 2 \bar{c}+\bar{a}$ as coterminus edges is
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27 cubic units
Volume of parallelepiped
$$
\begin{array}{l}
=(2 \bar{a}+\bar{b}) \cdot[(2 \bar{b}+\bar{c}) \times(2 \bar{c}+\bar{a})] \\
=(2 \bar{a}+\bar{b}) \cdot[(4 \bar{b} \times \bar{c})+(2 \bar{b} \times \bar{a})+(2 \bar{c} \times \bar{c})+(\bar{c} \times \bar{a})] \\
=[8 \bar{a} \cdot(\bar{b} \times \bar{c})]+[4 \bar{a} \cdot(\bar{b} \times \bar{a})]+[2 \bar{a} \cdot(\bar{c} \times \bar{a})]+[4 \bar{b} \cdot(\bar{b} \times \bar{c})]+[2 \bar{b} \cdot(\bar{b} \times \bar{a})]+[\bar{b} \cdot(\bar{c} \times \bar{a})] \\
=[8 \bar{a} \cdot(\bar{b} \times \bar{c})]+0+[\bar{b} \cdot(\bar{c} \times \bar{a})] \\
=8[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{c})]=9 \bar{a} \cdot(\bar{b} \times \bar{c}) \\
=9[\bar{a} \quad \bar{b} \quad \bar{c}]=9(3)=27
\end{array}
$$
$$
\begin{array}{l}
=(2 \bar{a}+\bar{b}) \cdot[(2 \bar{b}+\bar{c}) \times(2 \bar{c}+\bar{a})] \\
=(2 \bar{a}+\bar{b}) \cdot[(4 \bar{b} \times \bar{c})+(2 \bar{b} \times \bar{a})+(2 \bar{c} \times \bar{c})+(\bar{c} \times \bar{a})] \\
=[8 \bar{a} \cdot(\bar{b} \times \bar{c})]+[4 \bar{a} \cdot(\bar{b} \times \bar{a})]+[2 \bar{a} \cdot(\bar{c} \times \bar{a})]+[4 \bar{b} \cdot(\bar{b} \times \bar{c})]+[2 \bar{b} \cdot(\bar{b} \times \bar{a})]+[\bar{b} \cdot(\bar{c} \times \bar{a})] \\
=[8 \bar{a} \cdot(\bar{b} \times \bar{c})]+0+[\bar{b} \cdot(\bar{c} \times \bar{a})] \\
=8[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{c})]=9 \bar{a} \cdot(\bar{b} \times \bar{c}) \\
=9[\bar{a} \quad \bar{b} \quad \bar{c}]=9(3)=27
\end{array}
$$
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