My approach to this task is to find a matrix $M_{S}(F)$ where $S= \{(3,2,2),(1,2,1),(2,3,2)\}.$
But I have also thought that is keep staying in my mind that we are not really given a basis so it might be impossible to find a kernel as it is not said if there are no other elements for (example $(a,b,c)$ for which $F(a,b,c) = (c,d,e)$ etc.).
So is it possible to find a $\operatorname{kernel}(F)$ or not hmm?
I guess the problem here is to check whether $F$ is well-defined in the first place.
Normally, if a linear function (in particular a linear operator) is defined on a basis, it uniquely gets extended to the whole domain. In our case, it is defined on the vectors $(3,2,2), (1,2,1), (2,3,2)$, so all it takes is to check if those vectors form a basis. Depending on what you know about vector spaces, there are different ways of doing that: one possible way is to turn those vectors into columns and check if the matrix that you obtain, i.e. $\begin{bmatrix}3&1&2\\2&2&3\\2&1&2\end{bmatrix}$, is non-singular (i.e. has a nonzero determinant). Indeed, its determinant is $1$, which proves that $(3,2,2), (1,2,1), (2,3,2)$ is a basis.
Now we know $F$ is a well-defined linear operator, you can proceed to find its kernel.