How to prove there exists an isometry mapping all but the first two coordinates of $x, y$ and $z$ to zero 
For any $x, y, z\in\mathbb R^n(n>2)$, there exists an isometry mapping all but the first two coordinates of $x, y$ and $z$ to zero.

How to prove the statement above? I know that translation and rotation are isometries, but I’m just wondering if we can formulate a strict proof.
I tried in the following way but got stuck.
Let $L_{*\star}$ be the distance between $*$ and $\star$ and let $(x_1’,x_2’,0,\ldots,0), (y_1’,y_2’,0,\ldots,0), (z_1’,z_2’,0,\ldots,0)$ be the corresponding points. Since it is a distance-preserving mapping, we have
$$\left\{\begin{array}{l}
L_{xy}=\sqrt{(x_1’-y_1’)^2+(x_2’-y_2’)^2}\\
L_{xz}=\sqrt{(x_1’-z_1’)^2+(x_2’-z_2’)^2}\\
L_{yz}=\sqrt{(y_1’-z_1’)^2+(y_2’-z_2’)^2}
\end{array}\right.$$
I got stuck when trying to prove that system of equations has at least one solution. I suspect I’m doing it in the wrong way. Could you give me some help? Thank you!
 A: Given points $x,y,z\in \mathbb{R}^n$, we can define an isometry $M:\mathbb{R}^n\rightarrow \mathbb{R}^n$ by $M(w) = T(w-z)$ where $T$ is an orthogonal matrix.
General idea
Consider the $n\times 2$ matrix with column vectors $x-z$, $y-z$.  Apply rotators or reflectors as in QR decomposition to construct the matrix $T$.
Example
Let $x=[2,3,3,1]^\top$, $y=[2,3,5,2]^\top$, $z=[1,1,1,1]^\top$
Initially we form the matrix
\begin{equation}
[x-z \ y-z]
=
\left[
\begin{matrix}
1 & 1\\
2 & 2\\
2 & 4\\
0 & 1
\end{matrix}
\right]
\end{equation}
The QR decomposition of this matrix is (numerically with floating point precision)
\begin{equation}
\left[
\begin{matrix}
-0.3333333 &  -0.24759378&  -0.89364759& -0.17023569\\
-0.66666667 & -0.49518757 &  0.29595756 &  0.47196817\\
-0.66666667&  0.61898446 &  0.15086624& -0.38685032\\
-0.        &  0.55708601& -0.30173248 &  0.77370064
\end{matrix}
\right]
\left[
\begin{matrix}
-3.        & -4.33333333\\
 0.        &  1.79505494 \\
 0.        &  0.        \\
 0.        &  0.       
\end{matrix}
\right]
\end{equation}
Let $T$ be the transpose of the matrix $Q$ which is orthogonal. Then $M$ will map $x$ to the first column of $R$, $y$ to the second column of $R$, and $z$ to 0.
