Area of Parallelogram with two adjacent 2d vectors.

The area is equal to $\left|\mathbf{u} \times \mathbf{v}\right|$.

Notice that $\mathbf{u}\times \mathbf{v} = \det\begin{bmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&-2&0\\6&-2&0\end{bmatrix} = \det\begin{bmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&-2&0\\1&0&0\end{bmatrix} = 2\mathbf{k}$

Therefore $\left|\mathbf{u} \times \mathbf{v}\right| = |2\mathbf{k}| = 2$

Therefore the area of the required parallelogram is 2 sq. units.