b \times \bar c)=D $$

So,

$$ |D|=3 $$

The volume of a tetrahedron whose coterminous edges are three vectors $\bar u,\bar v,\bar w$ is

$$ \frac{1}{6}|\bar u \cdot (\bar v \times \bar w)| $$

Here the three edges are

$$ \bar a \times \bar b,\quad \bar a \times 2\bar c,\quad \bar b \times 2\bar c $$

Now,

$$ \bar a \times 2\bar c=2(\bar a \times \bar c) $$

and

$$ \bar b \times 2\bar c=2(\bar b \times \bar c) $$

So the required scalar triple product is

$$ (\bar a \times \bar b)\cdot\{2(\bar a \times \bar c)\times 2(\bar b \times \bar c)\} $$

$$ =4(\bar a \times \bar b)\cdot\{(\bar a \times \bar c)\times(\bar b \times \bar c)\} $$

Using vector product formula,

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

Now,

$$ (\bar a \times \bar c)\cdot \bar c=0 $$

and

$$ (\bar a \times \bar c)\cdot \bar b=-D $$

So,

$$ (\bar a \times \bar c)\times(\bar b \times \bar c)=D\bar c $$

Hence,

$$ 4(\bar a \times \bar b)\cdot(D\bar c) =4D\{(\bar a \times \bar b)\cdot \bar c\} $$

But

$$ (\bar a \times \bar b)\cdot \bar c=D $$

So the scalar triple product becomes

$$ 4D^2 $$

Since $|D|=3$,

$$ 4D^2=4(3)^2=36 $$

Volume of the tetrahedron is

$$ \frac{1}{6}\times 36=6 $$

So the correct option is

$$ \boxed{6} $$

Option A.

Vector Algebra

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