Least squares problem: am I solving it correctly? So I have this question:

In $\mathbb R^3$ with inner product calculate all the least square solutions, and choose the one with shorter length, of the system:
  $ x + y + z = 1 $
  $ x + z = 0 $
  $ y = 0 $

My attempt: So I applied the formula $ A^T A x = A^T b $ with A as being the matrix with row 1 (1,1,1) row 2 (1,0,1) and row 3 (0,1,0); x being the column $(x_1,x_2,x_3)$ and b being the column (1,0,0). 
So I did it and I reached to the solution $(x_1, \frac {1}{3}, \frac {1}{3} + x_1)$
And I expanded this solution in two vectors $(0, \frac {1}{3}, \frac {1}{3}) $ and $(1,0,1)$. 
So these are the least square solutions and the one with shorter length is the first one. 
My doubt is if I'm doing this correctly or if I made any mistake because I used an online calculator that only give one least square solution. Can someone help me to verify my attempt? Thanks!
 A: The reduced row echelon form is
$$
\begin{align}
  \mathbf{A} &\mapsto \mathbf{E}_{\mathbf{A}} \\
%
\left[
\begin{array}{ccc}
 1 & 1 & 1 \\
 1 & 0 & 1 \\
 0 & 1 & 0 \\
\end{array}
\right]
%
&\mapsto
%
\left[
\begin{array}{ccc}
 1 & 0 & 1 \\
 0 & 1 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
%
\end{align}
$$
There is a rank defect of $1$. Row 1 = row 2 + row 3.
The linear system is
$$
\begin{align}
  \mathbf{A} x &= b \\
%
\left[
\begin{array}{ccc}
 1 & 1 & 1 \\
 1 & 0 & 1 \\
 0 & 1 & 0 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 x \\
 y \\
 z
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
 1 \\
 0 \\
 0
\end{array}
\right]
%
\end{align}
$$
The least squares solution is resolved to $\color{blue}{range}$ and $\color{red}{null}$ space components
$$
\begin{align}
  x_{LS} &= 
\color{blue}{\mathbf{A}^{\dagger}b} + 
\color{red}{\left( \mathbf{I}_{3} - \mathbf{A}^{\dagger} \mathbf{A} \right) \xi} \\
%
&=
%
\color{blue}{\frac{1}{6}
\left[
\begin{array}{crr}
 1 & 2 & -1 \\
 2 & -2 & 4 \\
 1 & 2 & -1 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 1 \\
 0 \\
 0
\end{array}
\right]}
%
+ 
%
\color{red}{
\frac{1}{2}
\left[
\begin{array}{rcr}
 1 & 0 & -1 \\
 0 & 0 & 0 \\
 -1 & 0 & 1 \\
\end{array}
\right] \xi}, \quad \xi \in \mathbf{C}^{n}  \\[3pt]
%
&=
\color{blue}{
 \frac{1}{6}
\left[
\begin{array}{r}
  1 \\
  2 \\
  1
\end{array}
\right]}
+
\color{red}{
 \alpha
\left[
\begin{array}{r}
 -1 \\
  0 \\
  1
\end{array}
\right]}, \quad \alpha \in \mathbb{C}
%
\end{align}
$$
where $\xi$ is an arbitrary vector in the domain $\mathbb{C}^{3}$.
The residual error vector is
$$
\color{red}{r} =   \mathbf{A} \color{blue}{x_{LS}} - b = \color{red}{\frac{1}{3}
\left[
\begin{array}{r}
 -1 \\
  1 \\
  0
\end{array}
\right]},
$$
and the total error is $r^{2} = \frac{2}{9}.$
