Can three points from a 2-dimensional plane always be transformed to an equilateral triangle in 3-dimensional space? Given the following three points
$$A(x_{1},y_{1})\\
B(x_{2},y_{2})\\
C(x_{3},y_{3})\\$$
and assuming that at least two of these given points are different, how can $z_{1}$, $z_{2}$ and $z_{3}$ be defined so that
$$A'(x_{1},y_{1},z_{1})\\
B'(x_{2},y_{2},z_{2})\\
C'(x_{3},y_{3},z_{3})\\$$
form an equilateral triangle?  
It's easy to follow intuitively that such $z$ values always exist as long as at least two of the original points are not equal, but how could this be proven rigorously?
 A: What you have is a system of $3$ (non-linear) equations:
$$d(A',B')=d(A',C')=d(B',C')$$
where $d(\cdot,\cdot)$ is the usual Euclidean distance. This system has $3$ variables $z_1,z_2,z_3$, see if you can show that a solution exists.
A: You can assume without loss of generality that one of your points is $(0,0,0)$ and the other two points are $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$. Then you want to show you can choose $z_1,z_2$ so that $x_1^2 + y_1^2 + z_1^2 = x_2^2 + y_2^2 + z_2^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2$. The first two equations can be satisfied for some $z_2$ given a value for $z_1$, assuming $x_1^2 + y_1^2 \geq x_2^2 + y_2^2$. Furthermore, you will find that you can tune the values of $z_1,z_2$ so that $(z_1 - z_2)^2$ is within a wide range, either taking $z_1,z_2$ to have the same sign or opposite signs , and by considering $z_1$ as a free parameter you thus want to show you can make the third expression equal to the first two, which will probably be easiest if you use the intermediate value theorem from calculus.
