(i) You are sort of on the right track, but you are trying to (1) prove too much; and (2) drawing the wrong conclusion.
If you have a bunch of linearly independent vectors, then you can always find (at least one) linear transformation that maps them to whatever you want. This because we can always do it with a basis (decide where you want the basis vectors to go, and that gives you a linear transformation), and we can always take any collection of linearly independent vectors and complete it to a basis.
If your three vectors were linearly independent (hence a basis for $\mathbb{R}^3$, then you would know that there is a linear transformation with the desired values, because you can always find a linear transformation that does whatever you want to a basis.
However, if the three vectors are not linearly independent, this does not preclude the possibility that there is a linear transformation with the desired properties! For instance, $(1,0,0)$, $(0,1,0)$, and $(1,1,0)$ do not form a basis for $\mathbb{R}^3$, but I can nonetheless find a linear transformation such that $f(1,0,0)=(2,1,-1)$, $f(0,1,0) = (0,1,7)$, and $f(1,1,0)=(2,2,6)$ (in fact, many); for instance, $f(x,y,z) = (2x,x+y,7y-x)$ will do it.
The problem with your argument is that you misusing statement. You know that "If they are a basis, then a linear transformation exists". You are trying to use this by saying "If they are not a basis, then there is no linear transformation", but that is false. What you can conclude is that "if there is no linear transformation, then they are not a basis" (this is called the "contrapositive"). In general, if you know that "If P then Q", then from "P is false" you cannot conclude anything. Q may or may not hold. For instance, "If I fall into the pool, I will get wet." Does that mean that if I don't fall into the pool, I will be dry? No; I could get caught in a rainshower. If I know I didn't fall into the pool, I just don't know whether I will be wet or dry.
For this problem, thoguh, you have actually put your finger on exactly what you need to put your finger on: the fact that one of your vectors is a linear combination of the other two. Since $(2,0,1) = 2(1,1,1)-(0,2,1)$, that means that for any linear transformation $T$, you must have
$$T(2,0,1) = T\bigl(2(1,1,1)-(0,2,1)\bigr) = 2T(1,1,1) - T(0,2,1).$$
So you would need $f(2,0,1)=(1,0,0)$ to be equal to $$2f(1,1,1)-f(0,2,1)=2(3,2,7)-(2,1,-1) = (6-2, 4-1, 14+1) = (4,3,15).$$
Since $(1,0,0)\neq (4,3,15)$, there can be no linear transformation with the desired properties.
Added: The key here is that if you have vectors that satisfy some linear dependence, then any linear transformation must also respect that linear dependence. Since $(2,0,1) = 2(1,1,1)-(0,2,1)$, then any linear transformation $f$ will necessarily satisfy that $f(2,0,1) = 2f(1,1,1)-f(0,2,1)$. Since the values you have in (a) don't satisfy this, then you know there is no such linear transformation.
What if they do satisfy it, as in (b)? Here you have that $(0,2,-1)=2(1,0,2)+(-2,2,-5)$, so if $f$ is a linear transformation, then you must have
$$f(0,2,-1) = 2f(1,0,2) + f(-2,2,-5).$$
Here, this does happen, since
$$2f(1,0,2) + f(-2,2,-5) = 2(0,-2,1) + (3,3,1) = (3, -1, 3) = f(0,2,-1).$$
So, since they satisfy the equations, does there exist a linear transformation that does this? Yes! In fact, there are infinitely many different linear transformations that do this.
How can you find one? Well, this is where you can try doing systems of linear equations. You want to find a linear transformation
$$f(x,y,z) = (ax+by+cz, \ell x+my+nz, rx+sy+tz)$$
such that if you plug in $x=0$, $y=2$, and $z=-1$, you get $(3,-1,3)$; and if you plug in $x=1$, $y=0$, $z=2$, you get $(0,-2,1)$. This gives you six equations in nine unknowns (the unknowns being $a$, $b$, $c$, $\ell$, $m$, $n$, $r$, $s$, and $t$), and it will have a solution. (The other equation you get when you plug in $x=-2$, $y=2$, and $z=-5$ is actually a consequence of the six we already had; taht is what we checked to begin with). So you can try solving that system and getting one possible map.
(When the three vectors you have are a basis, you will get nine equations in nine unknowns with a single possible solution, so that will be the linear transformation you want.)
But there is a way here to save yourself a lot of work: take your favorite two vectors from this set of three, and then find a fourth vector that, together with the first two, makes up a basis. Then define the linear transformation by picking an arbitrary value for the image of that one.
For instance, start with $(0,2,-1)$ and $(1,0,2)$. Notice that $(1,0,0)$ is not a linear combination of these two, so $(0,2,-1)$, $(1,0,2)$, and $(1,0,0)$ makes a basis. Since we can always find a linear transformation that maps $(0,2,-1)$ to $(3,-1,3)$, $(1,0,2)$ to $(0,-2,1)$, and $(1,0,0)$ to anything (say $(0,0,0)$ because it makes things easier), that one will do. Notice that, if $f(1,0,0) = (0,0,0)$, that tells you that $a=\ell=r=0$, so now we only have three more unknowns. In fact, if we can figure out what $f(0,1,0)$ must be, that will gives us automatically $b$, $m$, and $s$; and $f(0,0,1)$ will give us $c$, $n$, and $t$. Can we do that? Yes: because $(0,2,-1)$, $(1,0,2)$, and $(1,0,0)$ are a basis: we have
$$(0,1,0) = \frac{1}{2}(0,2,-1) + \frac{1}{4}(1,0,2) -\frac{1}{4}(1,0,0).$$
So
$$f(0,1,0) = \frac{1}{2}(3,-1,3) + \frac{1}{4}(0,-2,1) - \frac{1}{4}(0,0,0).$$
Then to find the values of $c$, $n$, and $t$, we use the fact that
$$(0,0,1) = \frac{1}{2}(1,0,2) - \frac{1}{2}(1,0,0).$$
This also works when you are trying to find your function $f$ explicitly when the three vectors you have are a basis: try to write $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ as linear combinations of the three vectors (this amounts to solving three systems of three linear equations with three unknowns first; or to finding the inverse of a matrix; so it looks like the same amount of work, but because there are a lot of zeros it is often easier to solve these than the general one with $a$, $b,\ldots,t$). And then use the value that $f$ must have at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ to figure out the "formula" for $f$.