$$\left(\begin{smallmatrix}1&0&-2&~~~&0\\0&1&0&&0\\0&0&0&&0\end{smallmatrix}\right)\to \mathbb{L}=\langle\left(\begin{smallmatrix}2\\0\\1\end{smallmatrix}\right)\rangle$$
I have seen a lot ot tutorials, but none explaining how to assemble the solution $\mathbb{L}$, when $(0, 0, 0)$ is the solution of the whole matrix.
In other words, how can $(2, 0, 1)$ as a solution be possible?
On
You're trying to solve the matrix equation
$$\left(\begin{matrix} 1 & 0 & -2\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{matrix}\right) \left(\begin{matrix} x \\ y \\ z \end{matrix}\right) =\left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right) $$
Subsituting $$\left(\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}\right)$$ for $$\left(\begin{matrix} x \\ y \\ x \end{matrix}\right)$$ will verify that it is a solution.
The augmented matrix on the left is equivalent to $$\left(\begin{smallmatrix}1&0&-2\\0&1&0\\0&0&0\end{smallmatrix}\right)\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)=\left(\begin{smallmatrix}0\\0\\0\end{smallmatrix}\right)$$ Which in turn is equivalent to the system $$\{x-2z=0, y=0\}$$
This system has infinitely many solutions, but only a one-dimensional family $$\{(x,y,z):x=2z, y=0\}$$ This has basis $\{(2,0,1)\}$. However this isn't the only possible basis, another would be $\{(-6,0,-3)\}$. The key is that a basis must consist of exactly one vector that satisfies the system of equation. Why one? Because there are two pivots in the row echelon form, and hence $3-2=1$ free variable.