The area of the parallelogram whose adjacent sides are $\hat{i}-\hat{j}+3\hat{k}$ and $2\hat{i}-7\hat{j} + \hat{k}$ is :

$15\sqrt{2}$

The correct answer is Option (4) → $15\sqrt{2}$

area = $|\vec a×\vec b|$

$\vec a×\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&3\\2&-7&1\end{vmatrix}$

$=20\hat i+5\hat j-5\hat k$

$|\vec a×\vec b|=\sqrt{20^2+5^2+(-5)^2}=\sqrt{450}$

$=15\sqrt{2}$