How many different parallelograms can be drawn if given three co-ordinates in 3D Cartesian vector? By different, I mean the angles that each parallelograms make are different, the magnitude of the vectors that make each one are different, etc... 
I had this question on a test, where we have to find the angles inside the parallelogram. I got a different answer than my friend. I argued that this was possible because the fourth co-ordinates and your entire parallelogram depends on which way you interpret the shape, while she argued that there were many ways to draw, but they'll all give you the same parallelogram, with the same angles and magnitude, given the three co-ordinates.
 A: Well, clearly, in the general case, you don't have a single parallelogram out of your three points. 
Let's call $A,B,C$ the three points. 
Let's compare the three situations:


*

*$AB$ and $AC$ are sides, and $BC$ a diagonal $(1)$

*$AB$ and $BC$ are sides, and $AC$ a diagonal $(2)$

*$BC$ and $AC$ are sides, and $AB$ a diagonal $(3)$
Each case generates a different parallelogram, with differents side lenghts, and internal angles. 
$(1)$ is $ABA'C$ with angles ($\angle A, \angle B+\angle C)$
$(2)$ is $ABCB'$ with angles ($\angle A+\angle C, \angle B)$
$(3)$ is $AC'BC$ with angles ($\angle A+\angle B, \angle C)$
A: Let's just look at the case where the parallelogram is in the plane.  For simplicity, assume that the three given points are $A=(0,0)$, $B=(2,0)$, and $C=(1,1)$.
Big Assumption: You are allowed to determine the order of the points around the parallelogram.
If the points are in order $A$, $C$, $B$, then the fourth point could be $(-1,-1)$ which makes a square and side lengths $\sqrt{2}$.
If the points are in the order $B$, $A$, $C$, then the fourth point could be $(3,1)$.  This generates a parallelogram with side lengths $2$ and $\sqrt{2}$.  This is not a rectangle.
