Area of a parallelogram in $ \mathbb{R}3 $ Let $ A = (2, -3, 1), B = (5, -3, -1), C = (-2, -3, 5) $ and $ D = (1, -3, 3) $
I have the above situation. I know this: $$ A = \| \vec{AB} \times \vec{AD} \| $$ gives the area of the parallelogram.
My end components are: $ 0\hat{i} - 4\hat{j} + 0\hat{k} = -4\hat{j} $
The area I get is $4$ for this... but it's wrong and don't know where I'm going wrong.

The problem is the vertices form a parallelogram, but not in the order $ABCD$.
Note that $\vec{AB}  = (3,0,-2) = \vec{CD}$, and $\vec{AC}  = (-4,0,4) = \vec{BD}$.
However, $\vec{AD} = (-1,0,2) \neq (-7,0,6) = \vec{BC}$. Thus, $AB \parallel CD$ and $AC \parallel BD$. 
Hence, the vertices of the parallelogram in order are $ABDC$. So, the area is $\|\vec{AB} \times \vec{AC}\|$. 
