In order to solve "this" problem, i have to transform my corner-points from a 2D Space to my 3D Space.
But my two coordinate fields are only defined by their relation to each other. They have the same distance-vector's relations. (i guess)
Is this even solvable?
The simplest transformation between the two coordinate systems, and the one I suspect you’ve got in mind, is an affine map. It’s convenient to work in homogeneous coordinates, which allows this map to be represented by a $4\times3$ matrix $M$. I’ll use lower-case letters for points in the 2-D $u$-$v$ coordinate system and upper-case for points in the 3-D coordinate system to avoid ambiguity.
The constraints provided by the three point pairs can be captured in the “bulk” matrix equation $$M \begin{bmatrix}\mathbf p_1&\mathbf p_2&\mathbf p_3\\1&1&1\end{bmatrix}=\begin{bmatrix}\mathbf P_1&\mathbf P_2&\mathbf P_3\\1&1&1\end{bmatrix}$$ from which it immediately follows that $$M = \begin{bmatrix}\mathbf P_1&\mathbf P_2&\mathbf P_3\\1&1&1\end{bmatrix} \begin{bmatrix}\mathbf p_1&\mathbf p_2&\mathbf p_3\\1&1&1\end{bmatrix}^{-1}.$$ The inverse of the right-hand matrix exists as long as the three points are not colinear. (If they are, and the 3-D points are, too, there’s not a unique affine map, whereas if the 3-D points aren’t colinear, then there’s no affine map.) As a sanity check on your computation, the last row of $M$ should be $(0,0,1)$.†
The images of the corners of the 2-D unit square are then found by multiplying their homogeneous coordinate vectors by $M$, but the results will be simple combinations of the columns of $M$: $$(0,0) \mapsto M_3 \\ (1,0) \mapsto M_1+M_3 \\ (0,1) \mapsto M_2+M_3 \\ (1,1) \mapsto M_1+M_2+M_3.$$ You need to dehomogenize, of course, but if you’ve done this correctly, the last coordinate will always be $1$, so all you need to do to get the corresponding inhomogeneous Cartesian coordinates is to drop it.
† In fact, in practice you can drop the last row of $M$ so that you get dehomogenized coordinates directly when you multiply a homogeneous coordinate vector by $M$. If the last element of the homogeneous coordinate vector $\mathbf p$ is $1$, then so will be the last element of $M\mathbf p$, and dehomogenizing the result is a matter of dropping this $1$, as noted elsewhere.