The "kernel" of a linear transformation, f, is set of all v such that f(v)= 0.
Here, we must have $x_1+ 2x_2+ x_3= 0$, $2x_1+ x_3+ x_4= 0$, and $3x_1+ 2x_2+ 3x_3+ x_4= 0$. That is three equations in four unknowns so we would expect that there will be "one degree of freedom" or dimension 4- 3= 1. Specifically, we can solve the first equation for $x_3$: $x_3= -x_1- x_2$. The second equation then becomes $2x_1+ (-x_1- x_2)+ x_4= x_1- x_2+ x_4= 0$ and the third equation becomes $3x_1+ 2x_2+ 3(-x_1- x_2)+ x_4= -x_2+ x_4= 0$. We can then write $x_4= x_2$ so the $x_1- x_2+ x)2= x_1+ 2x_2= 0$. Then $x_1= -2x_2$, $x_4= x_2$, and $x_3= -x_1- x_2= 2x_2- x_2= x_2$. $(x_1, x_2, x_3, x_4)= (-2x_2, x_2, x_2, x_2)= x_2(-2, 1, 1, 1). Yes, the kernel is one-dimensional, spanned by the single vector (-2, 1, 1, 1).
Once we know that the kernel has dimension 1, we know that the image has dimension 4- 1= 3.
$T$ is a linear map from $\mathbf{R}^4 \to \mathbf{R}^3$ with transformation matrix $\mathbf{A}$.
Your kernel is correct. Multiplying your matrix $\textbf{A}$ with the vector $v:=(-2,1,1,0)^t$ gives $\vec 0$, so you easily see that your vector actually lies in the kernel.
Linear algebra tells you, that the kernel of a linear map is a subspace of the domain-vector space. So from the above calculation, you immediately see that the line spanned by $v$ lies inside the kernel. Possibly, it could be a plane a higher dimensional subspace. But since the matrix $\mathbf A$ and, equivalently, $\text{Reff}(\mathbf A)$ have $\text{rank}=3$, the kernel is $1$-dimensional (see rank-nullity). So it is exactly your line.
You get the image of $3\mathrm e_1 - 2 \mathrm e_2 - \mathrm 3_3$ by (left-)multiplication with $\textbf A$ ($3\mathrm e_1-2\mathrm e_2-\mathrm e_3$ lies in $\mathbf{R}^4$, the image point lies in $\mathbf{R}^3$). I guess you know how to multiply a matrix and a vector?
Abstractly, you can use that your transformation is linear to break up the term $T(3\mathrm e_1-2\mathrm e_2-\mathrm e_3)$. Then you only need to know what $\textbf{A}$ does with the standard-base-vectors. And $T$ maps the vector $\mathrm e_1=(1,0,0,0)^t$ to $(2,2,1)^t$, the first column of $\textbf A$.
Best Answer
You have a basis $(e_1,e_2,e_3)$ for $\mathbb{R}^3$. To find a linear map $f\in\mathcal{L}(\mathbb{R}^3)$ such that $\text{Im}f=S$ where $S$ is a subspace of $\mathbb{R}^3$, notice that $\text{Im}f=\text{span}(f(e_1),f(e_2),f(e_3))$. Notice also that $f$ is entirely determined if you know $f(e_1)$, $f(e_2)$ and $f(e_3)$. Hence simply find a basis for $S$ and try to choose $f(e_1)$, $f(e_2)$ and $f(e_3)$ in an appropriate way.