Let us compute the null space of \begin{equation*} C=\begin{bmatrix} -4 & 6 & 1 \\ -1 & 4 & 1 \\ 5 & 6 & 7 \\ 4 & 7 & 1 \end{bmatrix} \end{equation*} which we write as $\nsp{C}$. Translating Definition NSM, we simply desire to solve the homogeneous system $\homosystem{C}$. So we row-reduce the augmented matrix to obtain \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0 & 0 \\ 0 & \leading{1} & 0 & 0 \\ 0 & 0 & \leading{1} & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
There are no free variables in the homogeneous system represented by the row-reduced matrix, so there is only the trivial solution, the zero vector, $\zerovector$. So we can write the (trivial) solution set as \begin{equation*} \nsp{C}=\set{\zerovector}=\set{\colvector{0\\0\\0}} \end{equation*}