How to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ $B = \begin{bmatrix}
    1 & 2  & 1 \\
    1 & 3  & 1 \\
    1 & 4  & 1
\end{bmatrix}$ and I need to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ .
I know that $AB = A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix} \mid A\cdot\begin{bmatrix} 2\\3\\4 \end{bmatrix} \mid A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix} = 0 \mid 0 \mid 0$
Then I can conclude that (assume $A_1,...,A_n$ are columns of $A$):
1) $A_1 + A_2 + A_3 = 0$
2) $2A_1 + 3A_2 +4 A_3 = 0$
Meaning: $A_1 + 2A_2 +3A_3 = 0$
But now I got stuck... How should I continue from here?
 A: Just to simplify notation, instead of $AB=0$, let us have a look at the transpose $B^TA^T=0$. 
So we are looking at the subspace
$$W'=\{A\in M_{3\times3}; CA=0\}$$
where $C=B^T=
\begin{pmatrix}
  1 & 1 & 1 \\
  2 & 3 & 4 \\
  1 & 1 & 1 \\
\end{pmatrix}
$.
Let us denote by $\vec a_1$, $\vec a_2$, $\vec a_3$ the columns of matrix $A$.
We are looking at matrices such that
$$CA=C(\vec a_1,\vec a_2,\vec a_3)=(C\vec a_1,C\vec a_2,C\vec a_3)=0,$$
i.e.
\begin{align*}
C\vec a_1&=\vec 0\\
C\vec a_2&=\vec 0\\
C\vec a_3&=\vec 0
\end{align*}
(We are using how multiplication on the left operates on columns, see also: Intuition behind Matrix Multiplication.)
So we found out that the columns of the matrix $A\in W'$ have to fulfill the linear system $C\vec a=\vec 0$
$$\begin{pmatrix}
  1 & 1 & 1 \\
  2 & 3 & 4 \\
  1 & 1 & 1 \\
\end{pmatrix}
\begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$
By solving this system of equations we get that the solution are precisely multiples of $(1,-2,-1)^T$. This means that
$$W'=\{
\begin{pmatrix}
  a & b & c \\
-2a &-2b&-2c\\
  a & b & c 
\end{pmatrix}; a,b,c\in\mathbb R\}.
$$
(The space $W'$ consists of those matrices where each column is a multiple of this vector.)
Now to get back to the original question, we have to do the transpose. The space $W$ is equal to 
$$W=\{
\begin{pmatrix}
  a &-2a & a \\
  b &-2b & b \\
  c &-2c & c 
\end{pmatrix}
\}; a,b,c\in\mathbb R\}.$$
Dimension of this space is $3$ and as a basis we can choose the matrices
$$\begin{pmatrix}
  1 &-2 & 1 \\
  0 & 0 & 0 \\
  0 & 0 & 0 \\
\end{pmatrix},
\begin{pmatrix}
  0 & 0 & 0 \\
  1 &-2 & 1 \\
  0 & 0 & 0 \\
\end{pmatrix},
\begin{pmatrix}
  0 & 0 & 0 \\
  0 & 0 & 0 \\
  1 &-2 & 1 
\end{pmatrix}.$$
A: 
Then I can conclude that (assume $A_1,...,A_n$ are columns of $A$):
  1) $A_1 + A_2 + A_3 = 0$
  2) $2A_1 + 3A_2 +4 A_3 = 0$

Close! Instead of rows they should be columns because matrix multiplication would take the dot products of the row vectors of the first matrix with the column vectors of the second.
By elementary row operations, $(-2 eq. 1 + eq. 2)$ $A_2+2A_3=0 \Rightarrow A_2=-2A_3$
Substituting this relation into $eq. 1$ gives us $A_1-A_3=0 \Rightarrow A_1=A_3$.
So, we can create a matrix that respects the previous conditions:
$A =
\begin{bmatrix}
 a&b  &c \\ 
 -2a&-2b  &-2c \\ 
 a&b  &c 
\end{bmatrix}
$ and we want to find such $a,b,c$ that $AB=0$
$\begin{bmatrix}
 a&b  &c \\ 
 -2a&-2b  &-2c \\ 
 a&b  &c 
\end{bmatrix}\cdot \begin{bmatrix}
    1 & 2  & 1 \\
    1 & 3  & 1 \\
    1 & 4  & 1
\end{bmatrix}=\begin{bmatrix}
a+b+c&2a+3b+4c&a+b+c\\
-2(a+b+c)&-2(2a+3b+4c)&-2(a+b+c)\\
a+b+c&2a+3b+4c&a+b+c
\end{bmatrix}$
See how the system we are trying to solve: $a+b+c=0, \quad\!\! 2a+3b+4c=0$ is similar to the one above with $A_1,A_2,A_3$. So, $-b=2a=2c$.
The original matrix $A$ can now be expressed as $A =
\begin{bmatrix}
 t&-2t  &t \\ 
 -2t&4t &-2t \\ 
 t&-2t  &t 
\end{bmatrix}
$ where $t\in \mathbb{R}$
To construct a basis, notice that the matrix function $A(t)$ is of single variable. Thus, any multiple of \begin{bmatrix}
 1&-2  &1 \\ 
 -2&4 &-2 \\ 
 1&-2  &1 
\end{bmatrix} forms a basis for $W$.
$$W=span\begin{bmatrix}
 1&-2  &1 \\ 
 -2&4 &-2 \\ 
 1&-2  &1 \\
\end{bmatrix}=t\begin{bmatrix}
 1&-2  &1 \\ 
 -2&4 &-2 \\ 
 1&-2  &1 \\
\end{bmatrix}{\Huge{|}}t\in \mathbb{R}$$
