Answer
See below
Work Step by Step
Assume $x=(x_1,x_2,x_3)\\
y=(y_1,y_2,y_3)\\
c\in R$
$T(x+y)=T((x_1,x_2,x_3)+(y_1,y_2,y_3))\\
=T(x_1+y_1,x_2+y_2,x_3+y_3)\\
=(-3(x_3+y_3),2(x_1+y_1)-(x_2+y_2)+5(x_3+y_3))\\
=(-3x_3-3y_3,2x_1+2y_1-x_2-y_2+5x_3+5y_3)\\
=(-3x_3,2x_1-x_2+5x_3)+(-3y_3,2y_1-y_2+5y_3)\\
=T(x_1,x_2,x_3)+T(y_1,y_2,y_3)\\
=T(x)+T(y)$
$T(cx)=T(c(x_1,x_2,x_3))\\
=T(cx_1,cx_2,cx_3)\\
=(-3cx_3,2cx_1-cx_2+5cx_3)\\
=c(-3x_3,2x_1-x_2+5x_3)\\
=cT(x_1x_2,x_3)\\
=cT(x)$
Hence, $T$ is a linear transformation
Obtain: $Ker(T)=\{x \in R^3:T (x)=0\}\\
=\{(x,y,z) \in R^3: T(x,y,z)=0\}\\
=\{(z,y,z) \in R^3: (-3z,2x-y+5z)=(0,0)\\
\rightarrow z=0;y=2x$
then $Ker(T)=\{(x,2x,0)\}=\{x(1,2,0)\}=span \{(1,2,0)\}\\
\rightarrow \dim [Ker(T)]=1 \ne 0$
$T$ is not one-to-one
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim R^3\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)]=2$
$Rng(T) \subset R^2$ and $\dim R^2=2=\dim [Rng(T)]$
$T$ is onto
Basic for $Rng(T)$ is $\{(1,0),(0,1)$
