Finding a vector close to vector $\vec{b}$ using $A^T$ and $A\vec{x}=\vec{b}$ I'm having a hard time understanding the rest of the steps after $A^TA\vec{x}=A^T\vec{b}$ to find $\vec{x}$

Problem:
Find the vector in $W= span\ \left(\right.\
\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right]$ ,
$\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right]$ ,
$\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right]\ \left.\right)$
which is closest to the vector $\vec{b}=\left[ \begin{array}{c} 3 \\ 4 \\ 5 \\ 6 \end{array} \right]$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
(Note that $\vec{v_1},\vec{v_2},\vec{v_3}$ in $W$ are linearly independent)
Solution:
$$W=Col(A)\ \ \ , \ \ \ A=\left[ \begin{array}{ccc} 
\ 1 & 2 & -1 \\ 
\ 0 & 0 & 0 \\
\ 1 & -1 & 1 \\ 
\ 1 & 0 & 0 \end{array} \right]$$
$$------------$$
$$A\vec{x}=\vec{b}$$
$$A^TA\vec{x}=A^T\vec{b}$$
$$A^TA=\left[ \begin{array}{ccc} 
\ 3 & 1 & 0 \\ 
\ 1 & 5 & -3 \\
\ 0 & -3 & 2 \\ 
\end{array} \right]
\ \ ,\ \ 
A^T\vec{b}=\left[ \begin{array}{c} 
\ 14 \\ 
\ 1 \\
\ 2 \\ 
\end{array} \right]$$
$$(some\ term?)=\left[ \begin{array}{ccc} 
\ 3 & 1 & 0 \\ 
\ 1 & 5 & -3 \\
\ 0 & -3 & 2 \\ 
\end{array}\ \ \right|\ 
\left. \begin{array}{c} 
\ 14 \\ 
\ 1 \\
\ 2 \\ 
\end{array} \right]
=\left[ \begin{array}{ccc} 
\ 1 & 0 & 0 \\ 
\ 0 & 1 & 0 \\
\ 0 & 0 & 1 \\ 
\end{array}\ \ \right|\ 
\left. \begin{array}{c} 
\ 6 \\ 
\ -4 \\
\ -5 \\ 
\end{array} \right]\ \ ,\ \ 
\vec{x}=\left[ \begin{array}{c} 
\ 6 \\ 
\ -4 \\
\ -5 \\ 
\end{array} \right]$$
$$------------$$
$$\vec{v}_{answer}=W\cdot \vec{x} = 
\left[ \begin{array}{c} 
\ 3 \\ 
\ 0 \\
\ 5 \\
\ 6 \end{array} \right]$$
My Confusion:


*

*Is there a name for this theorem or property : $A^TA\vec{x}=A^T\vec{b}$

*What is the process of simplification to take "$A^TA\vec{x}=A^T\vec{b}$" and get $\vec{x}$ by itself? The solution I'm given doesn't explain this well. ($\ \vec{x}=\{A^TA\ |\ A^T\vec{b}\}$ What?!)

*How does the $\vec{v}_{answer}=\{3,0,5,6\}$ relate to the vectors $\vec{v_1},\vec{v_2},\vec{v_3},$ and $A\vec{x}=\vec{b}$ ? For instance is $\vec{v}_{answer}$ interchangeable with $\vec{v_1},\vec{v_2},\vec{v_3},$ or is $\vec{v}_{answer}$ only used to find $\vec{b}$, ect.?


First bullet point: This is called the normal equations.
