Answer
a basis for the nullspace of $A$ consists of the zero vector
$$\left[\begin{aligned} 0\\ 0\\ 0 \\ 0 \end{aligned}\right].$$
Work Step by Step
Given the matrix
$$
\left[ \begin {array}{cccc} 2&6&3&1\\ 2&1&0&-2
\\ 3&-2&1&1\\ 0&6&2&0\end {array}
\right]
.
$$
The reduced row echelon form is given by
$$
\left[ \begin {array}{cccc} 1&0&0&0\\ 0&1&0&0
\\ 0&0&1&0\\0&0&0&1\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned} x_{1} &=0
\\
x_{2} &=0\\
x_{3} &=0\\
x_{4} &=0\\
\end{aligned}.
$$
The solution of the above system is $x_1=0$,$x_2=0$, $x_3=0$ and $x_4=0$. This means that the solution space of $Ax = 0 $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2} \\x_3 \\x_4 \end{aligned}\right]= \left[\begin{aligned}0\\0\\0\\0 \end{aligned}\right] .$$
So, a basis for the nullspace of $A$ consists of the zero vector
$$\left[\begin{aligned} 0\\ 0\\ 0 \\ 0 \end{aligned}\right].$$
