About the Special Solutions Method in general, for solving Ax = 0 [GStrang, P140 3.2.1] 
● MIT Lec 7 Course Notes: Letting a different free variable equal 1 and setting the other free variables equal to zero gives us other vectors in the nullspace.
  ● P133: The nullspace consists of all combinations of the special solutions.
  ● 38:25 of Lecture 7 :  "If I set the free variable to 0 and solve for the pivot variables, I'll get all 0s. No progress." - Prof Strang

I accept the following general method for finding the nullspace and will exemplify with Ex 3.2.1: 
$A = \begin{bmatrix}
1 & 2 & 2 & 4 & 6 \\
1 & 2 & 3 & 6 & 9 \\
0 & 0 & 1 & 2 & 3
\end{bmatrix} \implies RREF(A) = \begin{bmatrix}
1 & 2 & 0 & 0 & 0 \\
 &  & 1 & 2 & 3 \\
 &  & \mathbf{0} &  & 
\end{bmatrix}$ Thus, $\mathbf{Ax = 0} \implies$
$x_1 = -2a_2 \\ 
x_2 = a_2 \\
x_3 = -2a_4 -3a_5 \\ 
x_4 = a_4 \\                
x_5 = a_5 
$ $\implies \mathbf{x} = a_2\begin{bmatrix}
-2  \\
1 \\
0 \\ 
0\\
0 \\
\end{bmatrix} + a_4\begin{bmatrix}
0  \\
0 \\
-2 \\ 
1 \\
0 \\
\end{bmatrix} +
a_5\begin{bmatrix}
0  \\
0 \\
-3 \\ 
0 \\
1 \\
\end{bmatrix}. $
$1.$ I can't pinpoint why, but I'm tentative about this method (in the grey box above): for each free variable (there're 3 here), this method sets $1$ for it and sets the other free variables to $0$. How and why does each free variable have one of these special solns? How and why does this function?
$2.$ Since $a_2, a_4, a_5$ are free, each can be any scalar. Say I select $a_2 = a_4 = a_5 = 1$.
Then  $x = (-2, 1, -5, 1, 1)$, which is one vector. But it's wrong to infer that $ \ker(A) = \{$ all scalar multiples of this one vector $\}$. How and Why? 
This question precedes rank, REF, $\mathbf{Ax = b}$, linear independence, span, basis, dimension, dimensions/theorems of the 4 subspaces,  Orthogonality, Determinants, eigenvalues and eigenvectors, and linear transformations. Please omit them from answers. 
 A: From $\begin{pmatrix}1&2&0&0&0\\ 0&0&1&2&3\end{pmatrix}$ no further calculation is required. First fill up the matrix in this way:
$$\begin{pmatrix}1&2&0&0&0\\
\color{red}0&\color{red}{-1}&\color{red}0&\color{red}0&\color{red}0\\
0&0&1&2&3\\
\color{red}0&\color{red}0&\color{red}0&\color{red}{-1}&\color{red}0\\
\color{red}0&\color{red}0&\color{red}0&\color{red}0&\color{red}{-1}
\end{pmatrix}.$$
Sharp eyes provided you'll see that 
$$\left\{\begin{pmatrix}\color{green}2\\ \color{green}{-1}\\ \color{green}0\\ \color{green}0\\ \color{green}0\end{pmatrix},\begin{pmatrix}\color{green}0\\ \color{green}0\\ \color{green}2\\ \color{green}{-1}\\ \color{green}0\end{pmatrix},\begin{pmatrix}\color{green}0\\ \color{green}0\\ \color{green}3\\ \color{green}0\\ \color{green}{-1}\end{pmatrix}\right\}$$
is a basis for the kernel of $A$.
A: The free variables are so called because you can give them whatever value you want and get a solution of your system just by setting the nonfree variables according to the equations.
What you want is to get linearly independent solution of $Ax=0$, but at a first stage this is not really important. What you want, at first, is to express in a convenient way the solutions. Since any solution can be expressed as
$$
\begin{bmatrix}
-2a_2\\
\color{red}a_{\color{red}2}\\
-2a_4-3a_5\\
\color{red}a_{\color{red}4}\\
\color{red}a_{\color{red}5}
\end{bmatrix}
= a_2\begin{bmatrix}
-2  \\
\color{red}1 \\
0 \\ 
\color{red}0 \\
\color{red}0
\end{bmatrix} + a_4\begin{bmatrix}
0  \\
\color{red}0 \\
-2 \\ 
\color{red}1 \\
\color{red}0 
\end{bmatrix} +
a_5\begin{bmatrix}
0  \\
\color{red}0 \\
-3 \\ 
\color{red}0 \\
\color{red}1 
\end{bmatrix}
$$
you can see those three vectors in action.
When you look at them from the point of view of linear independence and dimension, you realize those three vectors are a basis for the null space: the fact that there are three free variables intuitively says that the null space has dimension three and this intuition is confirmed by checking the definitions.

The procedure works in full generality. We can see it by using linear maps.
If $x_{i_1},x_{i_2},\dots,x_{i_k}$ are the free variables, we can define a map $f\colon \mathbb{C}^k\to \mathbb{C}^n$ by sending a vector in $\mathbb{C}^k$ to the one obtained with the corresponding values given to the free variables. In our example, just to show the idea, the map would be
$$
\begin{bmatrix}
\alpha_1\\
\alpha_2\\
\alpha_3
\end{bmatrix}\mapsto
\begin{bmatrix}
-2\alpha_1\\
\alpha_1\\
-2\alpha_2-3\alpha_3\\
\alpha_2\\
\alpha_3
\end{bmatrix}
$$
This linear map is injective, because its null space is clearly $\{0\}$ and its image is exactly the null space of $A$. Thus it induces an isomorphism between $\mathbb{C}^k$ and the null space of $A$, where $k$ is the number of free variables.
