Q.If A$(3,2,-1)$, B$(-2,2,-3)$, C$(3,5,-2)$, D$(-2,5,-4)$ then

$A(3,2,-1),B(-2,2,-3),C(3,5,-2),D(-2,5,-4)$. Checking the midpoint of $AD$ against the midpoint of $BC$:

These match, so $AD$ and $BC$ are the diagonals of the parallelogram -- meaning the vertices in order

around the parallelogram are $A,B,D,C$ (not $A,B,C,D$ as listed), confirmed by checking side $AB$ equals side

$CD$: $\overrightarrow{AB}=(-5,0,-2)$ and $\overrightarrow{CD}=D-C=(-5,0,-2)$ -- indeed equal. So $A,B,D,C$

are the vertices of a parallelogram, taken in that order.

Area: using adjacent sides $\overrightarrow{AB}=(-5,0,-2)$ and $\overrightarrow{AC}=(0,3,-1)$ from vertex

$A$: …

Test whether the midpoints of the two 'cross' diagonals (AD and BC, not the naive AC and BD) coincide, since with these four labelled points that is the pairing that actually bisects; confirm with a matching pair of equal …

Assuming without checking that the points in the order A,B,C,D as given automatically form the parallelogram in that cyclic order, when the …