Contradict the uniqueness of least-squares solution 
Find a least-squares solution of $Ax=b$ for 
  $A=\begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 \\
\end{pmatrix}
,b=\begin{pmatrix}
-3\\
-1\\
0\\
2\\
5\\
1
\end{pmatrix}$

My approach:
$$A^{T}A=\begin{pmatrix}
6 & 2 & 2 & 2 \\
2 & 2 & 0 & 0 \\
2 & 0 & 2 & 0 \\
2 & 0 & 0 & 2 
\end{pmatrix}$$
$$A^{T}b=\begin{pmatrix}
4\\
-4\\
2\\
6
\end{pmatrix}$$
Seems $A^{T}A$ is not invertible. The aumented matrix for $A^{T}A\hat{x}=A^{T}b$ is $$\begin{pmatrix}
6 & 2 & 2 & 2 & 4\\
2 & 2 & 0 & 0 & -4\\
2 & 0 & 2 & 0 & 2\\
2 & 0 & 0 & 2 &6
\end{pmatrix}\sim\begin{pmatrix}
1 & 0 & 0 & 1 & 3\\
0 & 1 & 0 & -1 & -5\\
0 & 0 & 1 & -1 & -2\\
0 & 0 & 0 & 0 & 0
\end{pmatrix}$$ Hence the general least-squares solution of $Ax=b$ has the form 
$$\hat{x}=\begin{pmatrix}3\\-5\\-2\\0\end{pmatrix}+t\begin{pmatrix}-1\\1\\1\\1\end{pmatrix}$$ 
But the orthogonal projection $\hat{x}$ is always unique then why that free variable come$?$ Eventually I can't interpret my outcome$!$ Can anyone explain this outcome Geometrically as well as why it contradict uniqueness please.
Update: So the columns of $A$ are not linearly independent hence free variable arrive but then $\color{red}{why}$ the projection of $b$ on the column space are so many$?$

Matrix $A$ is not full rank column (since $C_1=C_2+C_3+C_4$) and therefore we have one free parameter for the solution which is not unique.
Indeed recall that least square method corresponds to find solutons for $Ax=p$ where $p$ is the projection of $b$ onto $Col(A)$. Therefore since $A$ is not full rank column the result follows.
