Kernel of a 3×3 matrix and basis

Row reduction. $L_2\leftarrow L_2-2L_1$, $L_3\leftarrow L_3-L_1$:
$$\begin{pmatrix} 1 & 2 & -1 \\ 0 & -3 & 3 \\ 0 & -3 & 3 \end{pmatrix}.$$
$L_3\leftarrow L_3-L_2$:
$$\begin{pmatrix} 1 & 2 & -1 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{pmatrix}.$$
Rank is $2$, so $\dim\ker A=1$.

Solve. From $-3y+3z=0$, $y=z$. From $x+2y-z=0$, $x=-2y+z=-z$. So $(x,y,z)=(-z,z,z)=z(-1,1,1)$.

Conclusion. $\ker A=\operatorname{span}((-1,1,1))$, a line.