<s>[INST] 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. 
 [/INST] 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 $A$.
</s>