I think we are working over $\mathbb{R}$ here, or maybe $\mathbb{C}$? In any case, the kernel of $A$ is the solution set (it is a linear subspace of $\mathbb{R}^3$ / $\mathbb{C}^3$. I'll just assume you are working over the reals from now on) of the equation $Ax=0$, $x\in\mathbb{R}^3$. This requires no transformation matrix to compute.
But here you are only asked about the dimension of the image and kernel, respectively. Remember that the dimension formula (not sure if that's the right English term, correct me if it's wrong please) yields
$ \dim_K (\mathrm{Im}(f)) + \dim_K(\mathrm{Ker}(f))=\dim_K(V), $
where $f:V\to W$ is a $K$-linear mapping between $K$-vector spaces.
In your case, $A$ represents an $\mathbb{R}$-linear mapping $\mathbb{R}^3\to\mathbb{R}^3$, so $\dim_K(V)=\dim_{\mathbb{R}}(\mathbb{R}^3)=3$. So if you can compute the dimension of the image or kernel, you automatically get the other one by that formula above.
As you mentioned, the image is the span of the columns of $A$. You can easily see that the third column is a linear combination of the first two, and that the first two columns are linearly independent. This yields $\dim(\mathrm{Im}(A))=2$, hence $\dim (\mathrm{Ker}(A))=1$. (The image itself is then the span of the two vectors $x=(1,0,1)^{\top}, y=(0,-1,-1)^{\top}$ in $\mathbb{R}^3$, i.e. consists of all linear combinations of the form $\alpha x+\beta y$, where $\alpha,\beta\in\mathbb{R}$ are scalars.)