Basis of a kernel

Row reduction. $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3-R_1$:
$$\begin{pmatrix} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & -1 & -1 \end{pmatrix}.$$
Swapping $R_2$ and $R_3$:
$$\begin{pmatrix} 1 & 1 & 2 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \end{pmatrix}.$$
The rank is $2$.

Solving $Ax=0$. The second row gives $-y-z=0$, i.e. $y=-z$. The first row gives $x=-y-2z=z-2z=-z$. Hence
$$(x,y,z)=(-z,\,-z,\,z)=z\,(-1,-1,1).$$

Check. $A(-1,-1,1)^T=(-1-1+2,\ -2-2+4,\ -1+0+1)^T=(0,0,0)^T$.

Conclusion. $\ker f=\operatorname{span}\big((-1,-1,1)\big)$; a basis is $\big((-1,-1,1)\big)$ and $\dim\ker f=1$.