According to my book,

There exists a linear transformation $\displaystyle T:R^3 \rightarrow R^3$ that has $\displaystyle (1,-1,2),(2,1,-1),(1,2,-3) \in KerT$ and $\displaystyle (1,-1,2) \in ImT$

They explain that (1,-1,2),(2,1,-1),(1,2,-3) are dependent vectors.

My question is: If you are dealing with the standard basis (or any set of 3 independent vectors), does there exist a linear transformation where $\displaystyle T(1,0,0)=T(0,1,0)=T(0,0,1) =(0,0,0,)$ and have $\displaystyle (1,0,0) \in ImT $