There is a rectangle in 3-dimensional space. If I am only given the $x$ and $y$ coordinates, how do I find the $z$ coordinates? We are given the $x$ and $y$ coordinates of the four vertices of a rectangle.
How do I find the $z$-coordinates?
I know that this 3 dimensional rectangle must obey all the usual conditions of rectangle.
Opposite sides are equal.
The angles between two adjacent sides are 90 degrees.
 A: Suppose you have coordinates $A(x_A,y_A), B(x_B,y_B), C(x_C,y_C),D(x_D,y_D)$, taking them clockwise around the rectangle.
You will note that these probably don't look like a rectangle if drawn in the 2D plane. That's because you don't have the z-coordinates.
Let's assume that the z-coordinate of A is $0$.
Find the vectors $AB$ and $AD$ and make it so that $AB . AD =0$
$AB = \begin{bmatrix} x_B-x_A\\ y_B-y_A \\ z_B -0\end{bmatrix}$
$AD = \begin{bmatrix} x_D-x_A\\ y_D-y_A \\ z_D -0\end{bmatrix}$
$AB.AD=0 \Rightarrow \begin{bmatrix} x_B-x_A\\ y_B-y_A \\ z_B -0\end{bmatrix} .  \begin{bmatrix} x_D-x_A\\ y_D-y_A \\ z_D -0\end{bmatrix} = 0$
$(x_B-x_A)(x_D-x_A)+(y_B-y_A)(y_D-y_A)+z_Bz_D=0$
$z_Bz_D=-(x_B-x_A)(x_D-x_A)-(y_B-y_A)(y_D-y_A)$
You can choose any pair of values $z_B, z_D$ that make this true.
Now you have full sets of coordinates for A, B and D. The final stage is to find C.
Note that $AB = DC$ (parallel sides of a rectangle). So $OC = OD + DC = OD + AB$
In particular, $z_C= z_D + (z_B-z_A)$
Worked example
Consider these points: $A(3,3,z_A), B(10,5,Z_B), C(11,8,z_C), D(4,6,z_D)$
Let $z_A=0$.
$AB = \begin{bmatrix} 7\\ 2 \\ z_B \end{bmatrix}$
$AD = \begin{bmatrix} 1\\ 3 \\ z_D \end{bmatrix}$
$AB.AD=0 \Rightarrow 7+6+z_Bz_D=0$
$z_Bz_D=-13$
Choose $z_B=-1$ and $z_D=13$
We now have $A(3,3,0), B(10,5,-1), C(11,8,z_C), D(4,6,13)$
$DC = AB = \begin{bmatrix} 7\\ 2 \\ -1 \end{bmatrix}$
$OD \begin{bmatrix} 4\\ 6 \\ 13 \end{bmatrix}$
$OC = OD + DC = \begin{bmatrix} 7\\ 2 \\ -1 \end{bmatrix}+\begin{bmatrix} 4\\ 6 \\ 13 \end{bmatrix} = \begin{bmatrix} 11\\ 8 \\ 12 \end{bmatrix}$
Final set of points: $A(3,3,0), B(10,5,-1), C(11,8,12), D(4,6,13)$
A: (Rewritten after missing an entire degree of freedom!)
Let's say the vertices are in (clockwise or counterclockwise order) $\vec{A} = (x_A, y_A, z_A)$, $\vec{B} = (x_B, y_B, z_B)$, $\vec{C} = (x_C, y_C, z_C)$, and $\vec{D} = (x_D, y_D, z_D)$.
In order for the quadrilateral to be a parallelogram, pairs of opposing sides must be equal:
$$\left\lbrace\begin{aligned}
\vec{B} - \vec{A} &= \vec{C} - \vec{D} \\
\vec{D} - \vec{A} &= \vec{C} - \vec{B} \\
\end{aligned}\right. \quad \iff \left\lbrace\begin{aligned}
x_A - x_B + x_C - x_D &= 0 \\
y_A - y_B + y_C - y_D &= 0 \\
z_A - z_B + z_C - z_D &= 0 \\
\end{aligned}\right. \tag{1}\label{G1}$$
which means we can ignore $\vec{C}$ completely, because it is defined exactly by the three other points (via $\vec{C} = \vec{B} + \vec{D} - \vec{A}$), if it is a parallelogram.  (And rectangles and squares are always also parallelograms.)
First, check if the coordinates are valid for any rectangle that is perpendicular to the $z$ axis. If
$$(x_D - x_A)(x_B - x_A) + (y_D - y_A)(y_B - y_A) = 0 \tag{2a}\label{G2a}$$
then all real $z_A = z_B = z_C = z_D$ are valid solutions, as are $z_A = z_B \ne z_C = z_D$ and $z_A = z_D \ne z_B = z_C$.  (The latter are solutions because changing $z_A \to z_A + h$ and $z_B \to z_B + h$, or $z_A \to z_A + h$ and $z_D \to z_D + h$ is still a rectangle, increasing $h$ in magnitude just makes the rectangle taller or wider.)
For nontrivial solutions, we need all four angles of the parallelogram to be right angles:
$$\left\lbrace\begin{aligned}
\bigl( \vec{D} - \vec{A} \bigr) \cdot \bigl( \vec{B} - \vec{A} \bigr) &= 0 \\
\bigl( \vec{A} - \vec{B} \bigr) \cdot \bigl( \vec{C} - \vec{B} \bigr) &= 0 \\
\bigl( \vec{B} - \vec{C} \bigr) \cdot \bigl( \vec{D} - \vec{C} \bigr) &= 0 \\
\bigl( \vec{C} - \vec{D} \bigr) \cdot \bigl( \vec{A} - \vec{D} \bigr) &= 0 \\
\end{aligned}\right.$$
Because $\vec{C} = \vec{D} + \vec{B} - \vec{A}$, these all simplify to the same rule,
$$(x_B - x_A)(x_D - x_A) + (y_B - y_A)(y_D - y_A) + (z_B - z_A)(z_D - z_A) = 0$$
i.e.
$$(z_B - z_A)(z_D - z_A) = - (x_B - x_A)(x_D - x_A) - (y_B - y_A)(y_D - y_A)$$
which we can solve for $z_D$ (noting that we have already covered the $z_A = z_B$ cases in $\eqref{G2a}$):
$$z_D = z_A - \frac{ (x_B - x_A)(x_D - x_A) + (y_B - y_A)(y_D - y_A) }{z_B - z_A}, \quad z_A \ne z_B \tag{2b}\label{G2b}$$
meaning we can choose any pair of reals $z_A \ne z_B$, and $\eqref{G2b}$ and
$$z_C = z_D + z_B - z_A$$
will yield a valid rectangle.

Let's test this with
$$\vec{A} = \left[\begin{matrix} 3 \\ 3 \\ z_A \end{matrix}\right], \quad
\vec{B} = \left[\begin{matrix} 10 \\ 5 \\ z_B \end{matrix}\right], \quad
\vec{C} = \left[\begin{matrix} 11 \\ 8 \\ z_B + z_D - z_A \end{matrix}\right], \quad
\vec{D} = \left[\begin{matrix} 4 \\ 6 \\ z_D \end{matrix}\right]$$
(which yields $13 \ne 0$ for $\eqref{G2a}$, so $z_A=z_B=z_C=z_D$ are not valid rectangles).
Rule $\eqref{G2b}$ simplifies to
$$(z_D - z_A)(z_B - z_A) = -13 \quad \iff \quad z_D = z_A - \frac{13}{z_B - z_A}, z_A \ne z_B$$
Here are some randomly generated solutions:
$$\begin{array}{cc|cccc}
z_A & z_B & \vec{A} & \vec{B} & \vec{C} & \vec{D} \\
\hline
0 & 1 & (3, 3, 0) & (10, 5, 1) & (11, 8, -12) & (4, 6, -13) \\
1 & 0 & (3, 3, 1) & (10, 5, 0) & (11, 8, 13) & (4, 6, 14) \\
1 & -1 & (3, 3, 1) & (10, 5, -1) & (11, 8, 5.5) & (4, 6, 7.5) \\
-3 & 10 & (3, 3, -3) & (10, 5, 10) & (11, 8, 9) & (4, 6, -4) \\
\end{array}$$
