"Determine the vectorial subspace of the vectorial space R3 generated by:", without using matrices I'm struggling a bit with this exercise. 
$$a) \{(1, 0, 1),(0, 1, 0),(−2, 1, −2)\} \\
b) \{(1, 0, 1),(0, 1, 0),(−2, 1, −2),(−3, 4, −3)\} \\
c) \{(0, 1, 0),(−2, 1, −2)\} \\
d) \{(1, −1, 1),(1, 0, −1),(2, −1, 0)\}.$$
My professor does this without matrices. He does something like this for the first one:
$$(x,y,z) = \alpha(1,0,1) + \beta(0,1,0) + \gamma(-2,1,-2)$$
$$\begin{cases} x = \alpha - 3\gamma \\  y = \beta+\gamma \\ z = \alpha - 2\gamma \end{cases} \Leftrightarrow \begin{cases} x + 2\gamma = \alpha \\ - \\ z = x + 2\gamma-2\gamma \end{cases} \Leftrightarrow \begin{cases} x + 2\gamma = \alpha \\ y = \beta + \gamma \\ z = x \end{cases}$$
$$\{ (x,y,z) \in \mathbb{R}^3 : z = x \} = \{(x,y,z) \in \mathbb{R}^3:x,y \in \mathbb{R}\}$$
Can someone explain to me exactly what was done? This is similar to how you check if a system is linearly independent/dependent but with x,y,z instead of zero. I understand what a subspace is. My main question here exactly what was done (step by step) and why x and y are undefined/$\in \mathbb{R}$?
 A: In words, you are searching all the $(x,y,z)$ for which it exists $(\alpha,\beta,\gamma)$ such that:
$$
(x,y,z)=\alpha (1,0,1) + \beta (0,1,0) + \gamma (−2,1,−2)\ \ \ \ \ \ \ \ (Eq. 1)
$$ 
A natural approach is, given $(x,y,z)$, to try to express $(\alpha,\beta,\gamma)$ in term of $(x,y,z)$. That's what your professor wrote:
$\alpha-2\gamma=x, \beta+\gamma =y$, plus a mandatory relation $z=x$ (otherwise no solution exists).
The interpretation is that after having imposed $z=x$ there always exists $(\alpha,\beta,\gamma)$ such that (Eq. 1) is fulfilled: you have 2 equations $\alpha-2\gamma=x, \beta+\gamma =y$ for $3$ parameters $(\alpha,\beta,\gamma)$. For instance you can take $\gamma=0$. Then $\alpha=x$ and $\beta=y$. 
In summary your only restriction is $z=x$, by consequence your subspace is:
 $$\{ (x,y,z) \in \mathbb{R}^3 : z = x \} = \{(x,y,x) \in \mathbb{R}^3:x,y \in \mathbb{R}^2\}$$
as your professor wrote.

note: there is certainly a typo in your question, I think you wrote: $\{(x,y,z) \in \mathbb{R}^3:x,y \in \mathbb{R}\}$ instead of $\{(x,y,x) \in \mathbb{R}^3:x,y \in \mathbb{R}^2\}$
A: We got
$$\left\{ \matrix{x=&\alpha-2\gamma\\ y=&\beta+\gamma\\z=&x} \right.$$
where $\alpha,\beta,\gamma$ are arbitrary real numbers.
So, even by setting $\gamma=0$, we see that $x$ and $y$ can be arbitrary, independently from each other, and the only real constrain is $z=x$.
That is, the exact statement here is that

An arbitrary vector $(x,y,z)$ is in the span of $(1,0,1),\ (0,1,0),\ (-2,1,-2)$ if and only if $x=z$.

One direction is verified since all three spanning vectors satisfy this property, and so do their linear combinations. 
For the other direction, if $z=x$, so that our vector is $(x,y,x)$, then we simply have
$$(x,y,x)\ =\ x\cdot(1,0,1) + y\cdot(0,1,0)$$
so it is indeed in the span already of the first two vectors (note that the 3rd vector is also in there). 
