= Solution
The reduced row echelon form gives $u_1=2u_4/21$, $u_2=-3u_4/7$, and $u_3=-u_4/7$. Taking $u_4=21t$ gives every <vector> in the <null space> as $t(2,-9,-3,21)^\top$. Thus a <basis> for the kernel of the <linear map> is
$$
\boxed{\left\{\begin{pmatrix}2\\-9\\-3\\21\end{pmatrix}\right\}.}
$$
Direct multiplication gives $A(2,-9,-3,21)^\top=0$. The <rank-nullity theorem> also checks the count: <matrix rank> three plus <nullity> one equals the domain <dimension> four.
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