Least Squares solution for a symmetric singular matrix I want to solve this system by Least Squares method:$$\begin{pmatrix}1 & 2 & 3\\\ 2 & 3 & 4 \\\ 3 & 4 & 5 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\5\\-2\end{pmatrix} $$ This symmetric matrix is singular with one  eigenvalue $\lambda1 = 0$, so $\ A^t\cdot A$ is also singular and for this reason I cannot use the normal equation: $\hat x = (A^t\cdot A)^{-1}\cdot A^t\cdot b  $.
 So I performed Gauss-Jordan to the extended matrix to come with $$\begin{pmatrix}1 & 2 & 3\\\ 0 & 1 & 2 \\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\3\\-1\end{pmatrix} $$
Finally I solved the $\ 2x2$ system: $$\begin{pmatrix}1 & 2\\\ 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}1\\3\end{pmatrix} $$ taking into account that the best $\ \hat b\ $ is $\begin{pmatrix}1\\3\\0\end{pmatrix}$
The solution is then $\ \hat x = \begin{pmatrix}-5\\3\\0\end{pmatrix}$
Is this approach correct ? 
EDIT
Based on the book 'Lianear Algebra and its applications' from David Lay, I also include the Least Squares method he proposes: $(A^tA)\hat x=A^t b  $
$$A^t b =\begin{pmatrix}5\\9\\13\end{pmatrix}, A^tA = \begin{pmatrix}14 & 20 & 26 \\
20 & 29 & 38 \\
26 & 38 & 50\end{pmatrix}$$ 
The reduced echelon from the augmented is: 
$$ \begin{pmatrix}14 & 20 & 26 & 5 \\
20 & 29 & 38 & 9 \\
26 & 38 & 50 & 13 \end{pmatrix} \sim \begin{pmatrix}1 & 0 & -1 & -\frac{35}{6} \\
0 & 1 & 2 & \frac{13}{3} \\
0 & 0 & 0 & 0
\end{pmatrix} \Rightarrow \hat x = \begin{pmatrix}-\frac{35}{6} \\ \frac{13}{3} \\ 0 \end{pmatrix}$$ for the independent variable case that $z=\alpha ,  \alpha=0 $ 

Your solution is incorrect for the following reason. When you perform the Gauss-Jordan elimination, you transform the original system 
$$\tag{1}
Ax=b
$$ to another
$$\tag{2}
SAx=Sb.
$$
But the least squares solutions of (1) and (2) do not coincide in general.
Indeed, the least squares solution of (1) is $A^{+}b$, the least squares solution of (2) is $(SA)^{+}Sb$. If $A$ is invertible, then 
$$(SA)^{+}S=(SA)^{-1}S=A^{-1}=A^{+}$$
and everything is OK, but in general case $(SA)^{+}S\ne A^{+}$.
In your case, in particular,
$$
S=\left(\begin{array}{rrr}
       1       &       0    &          0    \\   
       2       &      -1    &          0    \\   
       1       &      -2    &          1    \\
\end{array}\right),\quad
(SA)^{+}S=\left(\begin{array}{rrr}
     -11/6    &        4/3  &          0 \\      
      -1/3    &       1/3   &         0    \\   
       7/6    &       -2/3  &          0  \\
\end{array}\right),
$$
$$
A^{+}=\left(\begin{array}{rrr}
     -13/12    &      -1/6     &       3/4 \\    
      -1/6     &       0       &       1/6   \\  
       3/4     &       1/6     &      -5/12\\ 
\end{array}\right).$$
You can calculate the pseudoinverse matrix by using the rank factorization:
$$
A=BC,\quad B=\left(\begin{array}{rr}
     1  &   3\\
     2  &   4\\
     3  &   5\\
\end{array}\right),\quad
C=\left(\begin{array}{rrr}
1&1/2&0\\
0&1/2&1
\end{array}\right)
$$
(this decomposition comes from the fact the second column of $A$ is the arithmetic mean of the remaining columns).
It remains only to calculate the pseudoinverse matrix 
$$
A^{+}=C^{+}B^{+}=C^T(CC^T)^{-1}(B^TB)^{-1}B^T
$$
and the least squares solution is $A^{+}b$.
