The area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}-7\hat{j}+\hat{k}$ is :

$15\sqrt{2}$

The correct answer is option (3) → $15\sqrt{2}$

area = $|\vec a×\vec b|$

$=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&3\\2&-7&1\end{vmatrix}=20\hat i+5\hat j-5\hat k$

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

$=\sqrt{20^2+5^2+5^2}=\sqrt{450}$

$=15\sqrt{2}$ sq. units