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?

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
$$\begin{pmatrix}
  1 &-2 & 1 \\
  0 & 0 & 0 \\
  0 & 0 & 0 \\
\end{pmatrix},
\begin{pmatrix}
  0 & 0 & 0 \\
  1 &-2 & 1 \\
  0 & 0 & 0 \\
\end{pmatrix},
\begin{pmatrix}
  0 & 0 & 0 \\
  0 & 0 & 0 \\
  1 &-2 & 1 
\end{pmatrix}.$$
