Let us compute the null space of \begin{equation*} A=\begin{bmatrix} 2 & -1 & 7 & -3 & -8 \\ 1 & 0 & 2 & 4 & 9 \\ 2 & 2 & -2 & -1 & 8 \end{bmatrix} \end{equation*} which we write as $\nsp{A}$. Translating Definition NSM, we simply desire to solve the homogeneous system $\homosystem{A}$. So we row-reduce the augmented matrix to obtain \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 2 & 0 & 1 & 0 \\ 0 & \leading{1} & -3 & 0 & 4 & 0 \\ 0 & 0 & 0 & \leading{1} & 2 & 0 \end{bmatrix} \end{equation*}
The variables (of the homogeneous system) $x_3$ and $x_5$ are free (since columns 1, 2 and 4 are pivot columns), so we arrange the equations represented by the matrix in reduced row-echelon form to \begin{align*} x_1&=-2x_3-x_5\\ x_2&=3x_3-4x_5\\ x_4&=-2x_5\\ \end{align*}
So we can write the infinite solution set as sets using column vectors, \begin{equation*} \nsp{A}=\setparts{ \colvector{-2x_3-x_5\\3x_3-4x_5\\x_3\\-2x_5\\x_5} }{ x_3,\,x_5\in\complex{\null} } \end{equation*}