Basis for the space of 2 by 3 matrices whose nullspace contains $(2,1,1)$ To find a basis for the space of $2$ by $3$ matrices whose nullspace contains $(2,1,1)$
My attempt:
Let A=   \begin{bmatrix}
        a & b & c \\
        d & e & f \\
        \end{bmatrix}
Then $(2,1,1)$ is a solution to $Ax=0$
So I get,
$$2a+b+c=0$$
$$2d+e+f=0$$
How will I proceed further?

You can use the Gram–Schmidt process to do this. For a certain $x_1$ and $x_2$ (such that $x$, $x_1$ and $x_2$ are linearly independent) you can get 
\begin{align}
y &=\begin{bmatrix}a \\ b \\ c\end{bmatrix}=x_1-x\cdot \frac{x\cdot x_1}{x \cdot x}, \\
z &=\begin{bmatrix}d \\ e \\ f\end{bmatrix}=x_2-x\cdot \frac{x\cdot x_1}{x \cdot x} -y\cdot \frac{y\cdot x_1}{y \cdot y}.
\end{align}
If you choose $x_1 = \begin{bmatrix} 0 & 6 & 0\end{bmatrix}^T$ and $x_2 = \begin{bmatrix} 0 & 0 & 30\end{bmatrix}^T$, you will end up with 
$$
\begin{bmatrix} y^T \\ z^T\end{bmatrix} = \begin{bmatrix}a & b & c \\ d & e & f\end{bmatrix} = \begin{bmatrix} -2 & 5 & -1 \\ -12 & 0 & 24\end{bmatrix}.
$$
In order to find a basis for the space of $2$ by $3$ matrices whose nullspace contains $\begin{bmatrix}2&1&1\end{bmatrix}$, I write the linear equations you mentioned in matrix format:
$$
\begin{bmatrix}2 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 1\end{bmatrix} 
\begin{bmatrix}a\\b\\c\\d\\e\\f\end{bmatrix}=
\begin{bmatrix}0\\0\end{bmatrix}.
$$
Now let
$$
B = \begin{bmatrix}2 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 1\end{bmatrix},
$$
then the solution to your problem is the nullspace of $B$. This can be easily found using the result above:
$$
\text{Null}(B) = \begin{bmatrix}-2 & -12 & 0 & 0 \\5 & 0 & 0 & 0\\-1 & 24 & 0 & 0 \\0 & 0 & -2 & -12 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & -1 & 24\end{bmatrix}.
$$
