If $\vec{p}=\hat{i}+\hat{j}-2 \hat{k}$ and $\vec{q}=2 \hat{i}+\hat{j}-\hat{k}$, then the area of parallelogram having diagonals $(\vec{p}+\vec{q})$ and $(\vec{p}-\vec{q})$ is

$\sqrt{11}$ sq. unit

$\vec{p}=\hat{i}+\hat{j}-2 \hat{k}, ~~\vec{q}=2 \hat{i}+\hat{j}-\hat{k}$

So, $\vec{p}+\vec{q}=3 \hat{i}+2 \hat{j}-3 \hat{k}, ~~\vec{p}-\vec{q}=-\hat{i}-\hat{k}$

so area of parallelogram = $\frac{\left|(\vec{p}+\vec{q}) \times (\vec{p}-\vec{q})\right|}{2} = A$

We have divided by 2 since we are using diagonals to compute area not adjacent sides.

so  $(\vec{p}+\vec{q}) \times (\vec{p}-\vec{q}) = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -3 \\ -1 & 0 & -1 \end{array}\right|=-2 \hat{i}+6 \hat{j}+2 \hat{k}$

So  $\frac{|-2 \hat{i}+6 \hat{j}+2 \hat{k}|}{2} = A$

$A = |-\hat{i} + 3\hat{j}+\hat{k}| = \sqrt{(-1)^2+3^2+1}=\sqrt{11}$ sq. units