Area of the parallelogram, whose adjacent sides are given by the vectors $\vec a =2\hat i-\hat j+5\hat k$ and $\vec b = 2\hat i+\hat j+ 2\hat k$, is:

$\sqrt{101}$

The correct answer is Option (2) → $\sqrt{101}$

Area of parallelogram,

Area = $|a×b|$

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

$=|-7\hat i+6\hat j+4\hat k|$

$|a×b|=\sqrt{49+36+16}$

$=\sqrt{101}$ sq. units