Answer
See below
Work Step by Step
Let $x=(x,y,z)\in R^3$
We have: $T(x)=ax\rightarrow T(x,y)=\begin{bmatrix}
1 & 3 & 2\\
2 & 6 & 5
\end{bmatrix}\begin{bmatrix}
x\\
y \\
z\end{bmatrix}=(x+3y+2z,2x+6y+5z)$
Obtain: $Ker (T)=\{x \in R^3: T(x)=0\}\\
=\{(x,y,z)\in R^3: T(x,y,z)=(0,0,0)\\
=\{(x,y,z)\in R^3:(x+3y+2z,2x+6y+5z)=(0,0)$
We have the system: $x+3y+2z=0\\
2x+6y+5z=0\rightarrow x=-3y$
Thus, $Ker (T)=\{(-3y,y,0):z \in R\}\\
=\{y(-3,1,0):y\in R\}\\
=span \{(-3,1,0)\}
\rightarrow \dim [Ker (T)]=1$
$Ker(T)$ is the line passing through the origin $(0,0,0)$ with the direction $(-3,1,0)$
Then, $Rng(T)=\{T(x):x \in R^3\}\\
=\{T(x,y,z):x,y,z \in R\}\\
=\{(x+3y+2z,2x+6y+5z):x,y,z \in R\}
=\{ x(1,2)+y(3,6)+z(2,5): x,y,z \in R\}
=span \{(1,2);(3,6);(2,5)\\\rightarrow Rng(T)=R^2\\
\rightarrow \dim [Rng(T)]=2$
Thus, $Rng(T)$ is the whole 2 dimensional space.
Verify a Rank Nullity Theorem:
$\dim [Ker (T)]+\dim[Rng(T)]=2+1=3=\dim R^3$
