Notice that $ker(F)=\left\langle (1, 0, -1), (0,1,2) \right\rangle$ (why?) and also notice that (1,0-1) and (0,1,2) are linearly independent therefore using the Dimension Theorem :
$$3=dim(\mathbb R^3)=dim(ker(F))+dim(Im(F))=2+dim(Im(F)) \Rightarrow dim(Im(F))=1$$ now let B={(0,1,1),(1, 0, -1), (0,1,2)} , B is L.I and $B \subseteq \mathbb R^3$ therefore B is basis of $\mathbb R^3$ (why?)
so the only thing left to do will be to write the matrix of F with respect to base basis B as: $$ \begin{pmatrix} |& | & | \\ [F((0,1,1))]^t & [F((1,0,1-1)]^t & [F((0,1,2))]^t \\ |& | & | \\ \end{pmatrix}$$ knowing that (1, 0, -1) and (0,1,2) are in the kernel. I hope that helps!