<s>[INST] minimizing vectors Could please check whether my solution is right? 
Q. $$A=\begin{pmatrix}
        1 & 1 \\
        1 & 1 \\
        0 & 0 \\
        \end{pmatrix}$$Find the set of vectors $x$ that minimize the value $$||
        A
x-
        \begin{pmatrix}
        1  \\
        2 \\
        3  \\
        \end{pmatrix}||$$
My solution. 
$Ax-(1,2,3)^T=0$ (inconsistent) → least square method 
$A^TAx=A^Tb$,                $b=(1,2,3)^T$ 
$$
\begin{pmatrix}
        2 & 2 \\
        2 & 2 \
  \end{pmatrix}\begin{pmatrix}
        x_1 \\
        x_2 \\
        \end{pmatrix}=\begin{pmatrix}
        3 \\
        3 \\
        \end{pmatrix}$$
$2x_1+2x_2=3$ → $x_2=-x_1+1.5$ 
Therefore the set of $x$ is {$ 
\begin{pmatrix}
        k \\
        -k+1.5 \\
        \end{pmatrix}$| k∈ℝ }
 [/INST] I would use a more "basic" method.
Writing the vector x as $\begin{pmatrix}x \\ y \end{pmatrix}$,  $Ax= \begin{pmatrix}1 & 1 \\ 1 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+ y \\ x+ y \\ 0 \end{pmatrix}$ so that $Ax- \begin{pmatrix}1 \\ 2 \\ 3 \end{pmatrix}= \begin{pmatrix}x+ y- 1 \\ x+ y- 2 \\ -3\end{pmatrix}$.
The norm of that is $\sqrt{(x+ y- 1)^2+ (x+ y- 2)^2+ 9}$.
Writing that as f(x,y), we have $f_x= (1/2)((x+ y- 1)^2+ (x+ y- 2)^2+ 9)^{-1/2}[(2(x+y-1)+ 2(x+ y- 2)]= 0$ and $f_y= (1/2)((x+ y- 1)^2+ (x+ y- 2)^2+ 9)^{-1/2}[(2(x+y-1)+ 2(x+ y- 2)]= 0$. Those both reduce to $2[(x+ y- 1)+ 2(x+ y- 2)]= 4x+ 4y- 6= 0$.  The function will be minimized at any point along the line $x+ y= \frac{3}{2}$.  That is, $y= \frac{3}{2}- x$ which is exactly what you have.
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