Answer
See below
Work Step by Step
With $v \in V \rightarrow v=av_1+bv_2+cv_3$
Obtain: $T(v)=T(av_1+bv_2+cv_3)\\
=aTv_1+bTv_2+cTv_3\\
=a(2w_1-w_2)+b(w_1-w_2)+c(w_1+2w_2)\\
=2aw_1-aw_2+bw_1-bw_2+cw_1+2cw_2\\
=(2a+b+c)w_1+(-a-b+2c)w_2$
$Ker (T)=\{v \in V: T(v)=0\}\\
=\{av_1+bv_2+cv_3:(2a+b+c)w_1+(-a-b+2c)w_2: a,b,c \in R$
We have the system:
$2a+b+c=0\\
-a-b+2c=0$
thus, $a=-3c\\b=5c$
Thus, $Ker (T)=\{(-3cv_1+5cv_2+cv_3):c \in R\}\\
=\{c(-3v_1+5v_2+v_3):c \in R\}\\
=span\{(-3v_1+5v_2+v_3)\}$
$\dim [Ker(T)]=1$
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim V\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)]=2$
Since $Rng(T) \subset W$ and $\dim W=2$then $Rng(T)]=W$
