I have been doing one exercise and I don't understand one step of the logic, so I would be grateful if you could explain. This is the exercise:
I will briefly explain up to which point I got: First I find out the solution space V, which is:
Then I input this into:
and I get this: 
This is where I got. Now it says in the solutions: Consequently, $\text{Range}(T)$ is a plane in $\mathbb{R}^3$ whose first coordinate is $0$, so the basis is: $\{ (0, 1, 0),(0, 0, 1) \}$
My question: How is it concluded that $\text{Range}(T)$ is a plane in $\mathbb{R}^3$, and why is the basis of it $\{(0, 1, 0),(0, 0, 1)\}$?
The range of $T$ is a subspace of $\mathbb{R}^3$ and it follows from your computations that the first coordinate of each element of the range is $0$. On the other hand, the range has dimension at least $2$. For instance, putting $r=1$ and $s=t=0$, you get that $(0,0,5)$ is in the range. And putting $r=s=0$ and $t=1$, you get that $(0,7,1)$ is in the range. But $(0,0,5)$ and $(0,7,1)$ are linearly independent. And the only subspace of $\mathbb{R}^3$ with dmension at least $2$ and such that the first coordinate of each element is $0$ is $\{(0,a,b)\,|\,a,b\in\mathbb{R}\}$. A basis of this space is $\{(0,1,0),(0,0,1)\}$.