lelogram, the diagonal from $O$ to $D$ is

$$ \overline{OD}=\overline{OB}+\overline{OC} $$

and the other diagonal is

$$ \overline{CB}=\overline{OB}-\overline{OC} $$

Given,

$$ \overline{OD}=\hat{i}+2\hat{j}+6\hat{k} $$

and

$$ \overline{CB}=-3\hat{i}-2\hat{k} $$

So,

$$ \overline{OB}+\overline{OC}=\hat{i}+2\hat{j}+6\hat{k} $$

$$ \overline{OB}-\overline{OC}=-3\hat{i}-2\hat{k} $$

Adding these two equations,

$$ 2\overline{OB}=(-2)\hat{i}+2\hat{j}+4\hat{k} $$

$$ \overline{OB}=-\hat{i}+\hat{j}+2\hat{k} $$

Now subtracting the second equation from the first,

$$ 2\overline{OC}=4\hat{i}+2\hat{j}+8\hat{k} $$

$$ \overline{OC}=2\hat{i}+\hat{j}+4\hat{k} $$

Also,

$$ \overline{OA}=\hat{i}+2\hat{j}+3\hat{k} $$

The volume of the parallelopiped formed by $\overline{OA}$, $\overline{OB}$ and $\overline{OC}$ is the magnitude of their scalar triple product:

$$ V=\left|\overline{OA}\cdot(\overline{OB}\times \overline{OC})\right| $$

Using determinant form,

$$ V= \left| \begin{vmatrix} 1 & 2 & 3\\ -1 & 1 & 2\\ 2 & 1 & 4 \end{vmatrix} \right| $$

Expanding,

$$ V=\left|1(4-2)-2(-4-4)+3(-1-2)\right| $$

$$ V=\left|2+16-9\right| $$

$$ V=9 $$

So, the volume is

$$ \boxed{9} $$

Correct option: Option C.

Vector Algebra

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