Find a vector which is perpendicular to the vectors $\vec{u}=\hat{j}+\sqrt{3}\hat{k}$ and $\vec{v}=\sqrt{3}\hat{j}+2\hat{k}$ whose magnitude is equal to the area of the parallelogram which is formed by $\vec{u}$ and $\vec{b}$.

The alternatives in my book are as follows:

$\begin{array}{ll}

1.&-2\hat{j}\\

2.&-\hat{i}\\

3.&3\hat{k}\\

4.&5\hat{i}\\

5.&3\hat{i}\\

\end{array}$

I'm totally lost at this question. What should I do to find the area? Does there exist a formula which can be used to relate it with the fact that a vector is perpendicular to those two? Can someone explain the solution step by step, so I can understand it?