Kernel and image of a matrix

Rank. The three rows are all multiples of $(1,2,3)$, so $\operatorname{rank}(A) = 1$.

Image. A basis of the image is the single nonzero column, i.e. $(1,2,1)$ (first column). So $\operatorname{Im} A = \operatorname{span}((1,2,1))$.

Kernel. Solve $Ax = 0$: $x_1 + 2x_2 + 3x_3 = 0$. The solutions are $(x_1,x_2,x_3) = (-2s-3t, s, t)$ with $s,t \in \mathbb{R}$. A basis is $(-2,1,0)$ and $(-3,0,1)$.

Conclusion. $\dim(\ker A) = 2$, $\dim(\operatorname{Im} A) = 1$, and $2+1=3$, consistent with the rank-nullity theorem.