ABCD is a parallelogram with $\overrightarrow{A C}=\hat{i}-2 \hat{j}+\hat{k}$ and $\overrightarrow{B D}=-\hat{i}+2 \hat{j}-5 \hat{k}$. Area of this parallelogram is equal to:

$2 \sqrt{5}$ sq. units

Area vector of parallelogram

$=\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{B D})$

$=\frac{1}{2}\left|\begin{array}{ccc}i & j & k \\ 1 & -2 & 1 \\ -1 & 2 & -5\end{array}\right|$

$=\frac{1}{2}(8 \hat{i}+4 \hat{j})$

$=4 \hat{i}+2 \hat{j}$

∴  Area of the parallelogram = $|4 \hat{i}+2 \hat{j}|=2 \sqrt{5}$ sq. units

Hence (2) is correct answer.