Answer
See below
Work Step by Step
Let $x=(x,y,z)\in R^3$
We have: $T(x)=ax\rightarrow T(x,y)=\begin{bmatrix}
1 & -2 & 1\\
2 & -3 & -1\\
5 & -8 & -1
\end{bmatrix}\begin{bmatrix}
x\\
y \\
z\end{bmatrix}=(x-2y+z,2x-3y-z,5x-8y-z)$
Obtain: $Ker (T)=\{x \in R^3: T(x)=0\}\\
=\{(x,y,z)\in R^3: T(x,y,z)=(0,0,0)\\
=\{(x,y,z)\in R^3:(x-2y+z,2x-3y-z,5x-8y-z)=(0,0,0)$
We have the system: $x-2y+z=0\\
2x-3y-z=0\\
5x-8y-z=0
\rightarrow x=5z,y=3z$
Thus, $Ker (T)=\{(5z,3z,z):z \in R\}\\
=\{z(5,3,1): z \in R\}\\
=span \{(5,3,1)\}
\rightarrow \dim [Ker (T)]=1$
$Ker(T)$ is the line passing through the origin $(0,0,0)$ with a direction $(5,3,1)$
Then, $Rng(T)=\{T(x):x \in R^3\}\\
=\{T(x,y,z):x,y,z \in R\}\\
=\{(x-2y+z,2x-3y-z,5x-8y-z):x,y,z \in R\}
=\{ x(1,2,5)+y(-2,-3,-8)+z(1,-1,1): x,y,z \in R\}
=span \{(1,2,5);(-2,-3,-8);(1,-1,1)\}$
Find a basic for $Rng(T)$
$\begin{bmatrix}
1 & -2 & 1\\
2 & -3 & -1 \\
5 & -8 & -1
\end{bmatrix}\approx\begin{bmatrix}
1 & 2 & 5\\
-3 & -3 & -8 \\
1 & -1 & 1
\end{bmatrix}\approx \begin{bmatrix}
1 & 2 & 5\\
0 & 1 & 2 \\
0 & -3 & -6
\end{bmatrix}\approx \begin{bmatrix}
1 & 2& 5\\
0 & 1 & 2 \\
0 & 0 & 0
\end{bmatrix} $
Hence, we have $\{(1,2,5);(0,1,2)\}$ is a basic for $Rng(T)$
$\rightarrow Rng(T)=span \{(1,2,5);(0,1,2)\}\\
\rightarrow \dim [Rng(T)]=2$
Thus, $Rng(T)$ is a plane containing points $(0,0,0);(1,2,5);(0,1,2)$
Verify a Rank Nullity Theorem:
$\dim [Ker (T)]+\dim[Rng(T)]=2+1=3=\dim R^3$
