Answer
See below
Work Step by Step
Let $x=(x,y,z)\in R^3$
We have: $T(x)=ax\rightarrow T(x,y)=\begin{bmatrix}
1 & -1 & 2\\
-3 & 3 & 6
\end{bmatrix}\begin{bmatrix}
x\\
y \\
z\end{bmatrix}=(x-y+2z,-3x+3y-6z)$
Obtain: $Ker (T)=\{x \in R^3: T(x)=0\}\\
=\{(x,y,z)\in R^3: T(x,y,z)=(0,0,0)\\
=\{(x,y,z)\in R^3:(x-y+2z,-3x+3y-6z)=(0,0)$
We have the system: $x-y+2z=0\\
-3x+3y-6z=0\rightarrow x=y-2z$
Thus, $Ker (T)=\{(y-2z,y,z):z \in R\}\\
=\{y(1,1,0)+z(-2,0,1):y, z \in R\}\\
=span \{(1,1,0);(-2,0,1)\}
\rightarrow \dim [Ker (T)]=2$
$Ker(T)$ is the plane containing $(0,0,0);(1,1,0)$ and $(-2,0,1)$
Then, $Rng(T)=\{T(x):x \in R^3\}\\
=\{T(x,y,z):x,y,z \in R\}\\
=\{(x-y+2z,-3x+3y-6z):x,y,z \in R\}
=\{ x(1,-3)+y(-1,3)+z(2,-6): x,y,z \in R\}
=span \{(1,-3);(-1,3);(2,-6)\\\rightarrow Rng(T)=span \{(1,-3)\}\\
\rightarrow \dim [Rng(T)]=1$
Thus, $Rng(T)$ is the line passing through the origin $(0,0)$ with the direction $(1,-3)$
Verify a Rank Nullity Theorem:
$\dim [Ker (T)]+\dim[Rng(T)]=2+1=3=\dim R^3$
