Answer
a basis for the nullspace of $A$ consists of the vectors
$$\left[\begin{aligned} 2\\ 1\\0 \end{aligned}\right], \left[\begin{aligned} -7\\ 0\\1 \end{aligned}\right].$$
Work Step by Step
Given the matrix
$$
A=\left[ \begin {array}{ccc} 3&-6&21\\ -2&4&-14
\\ 1&-2&7\end {array} \right]
.
$$
The reduced row echelon form is given by
$$
\left[ \begin {array}{ccc} 1&-2&7\\ 0&0&0
\\ 0&0&0\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned} x_{1} -2x_2+7x_3&=0
\\
\end{aligned}.
$$
The solution of the above system is $x_1=2s-7t$, $x_2=s$ and $x_3=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2} \\x_3 \end{aligned}\right]= \left[\begin{aligned} 2s-7t\\ s\\t \end{aligned}\right]=s\left[\begin{aligned} 2\\ 1\\0 \end{aligned}\right]+t\left[\begin{aligned} -7\\ 0\\1 \end{aligned}\right] .$$
So, a basis for the nullspace of $A$ consists of the vectors
$$\left[\begin{aligned} 2\\ 1\\0 \end{aligned}\right], \left[\begin{aligned} -7\\ 0\\1 \end{aligned}\right].$$
