The volume of a parallelopiped with coterminous edges $\bar{a}, \bar{b}, \bar{c}$ is 3 cubic units. The volume (in cubic units) of a tetrahedron with coterminous edges $(\bar{a} \times \bar{b}), (\bar{a} \times 2\bar{c}), (\bar{b} \times 2\bar{c})$ is...

The volume of the parallelopiped with coterminous edges $\bar{a}, \bar{b}, \bar{c}$ is given by $|[\bar{a} \bar{b} \bar{c}]| = 3$.

The volume of a tetrahedron with coterminous edges $\bar{x}, \bar{y}, \bar{z}$ is $V = \dfrac{1}{6} |[\bar{x} \bar{y} \bar{z}]|$.

Here, the edges are $\bar{x} = \bar{a} \times \bar{b}$, $\bar{y} = 2(\bar{a} \times \bar{c})$, and $\bar{z} = 2(\bar{b} \times \bar{c})$.

Evaluating the scalar triple product of these edges:

$[\bar{x} \bar{y} \bar{z}] = [\bar{a} \times \bar{b} \quad 2(\bar{a} \times \bar{c}) \quad 2(\bar{b} \times \bar{c})] = 4 [\bar{a} \times \bar{b} \quad \bar{a} \times \bar{c} \quad \bar{b} \times \bar{c}]$

Using the scalar triple product expansion:

$[\bar{a} \times \bar{b} \quad \bar{a} \times \bar{c} \quad \bar{b} \times \bar{c}] = (\bar{a} \times \bar{b}) \cdot ((\bar{a} \times \bar{c}) \times (\bar{b} \times \bar{c}))$

Applying the vector triple product formula:

$(\bar{a} \times \bar{c}) \times (\bar{b} \times \bar{c}) = [\bar{a} \bar{c} \bar{c}] \bar{b} - [\bar{a} \bar{c} \bar{b}] \bar{c}$

Since $[\bar{a} \bar{c} \bar{c}] = 0$ and $[\bar{a} \bar{c} \bar{b}] = -[\bar{a} \bar{b} \bar{c}]$, we get:

$(\bar{a} \times \bar{c}) \times (\bar{b} \times \bar{c}) = [\bar{a} \bar{b} \bar{c}] \bar{c}$

Substituting this back into the scalar triple product:

$(\bar{a} \times \bar{b}) \cdot ([\bar{a} \bar{b} \bar{c}] \bar{c}) = [\bar{a} \bar{b} \bar{c}] ((\bar{a} \times \bar{b}) \cdot \bar{c}) = [\bar{a} \bar{b} \bar{c}]^2$

Therefore, the scalar triple product of the new edges is:

$[\bar{x} \bar{y} \bar{z}] = 4 [\bar{a} \bar{b} \bar{c}]^2$

Since $|[\bar{a} \bar{b} \bar{c}]| = 3$, we have $[\bar{a} \bar{b} \bar{c}]^2 = 9$.

$[\bar{x} \bar{y} \bar{z}] = 4 \times 9 = 36$

The volume of the tetrahedron is $V = \dfrac{1}{6} \times 36 = 6$ cubic units.

Answer: $6$