I want to show that lines from this family in euclidean 3 dimensional space
$$\begin{cases} \alpha(x-y) + z = 0 \\ x+y+2 \alpha = 0 \end{cases}$$
do not intersect for different values of $\alpha$.
This is a family of lines because it's the intersection of two planes, I think they do not intersect because each line is perpendicular to the $z$ plane at the value of $ z = -\alpha(x-y) $ is this enough to conclude?
And this got me thinking how would I represent a family of lines all intersecting in one point in 3 dimensional euclidean space? Could I have a link or a source where to study the general derivation of parallel family of lines in 3 dimensional space?
With your last change to your question, we can answer it easily now. Take a point $(r,s,t)$ that satisfies your set of equations for both $\alpha_1$ and $\alpha_2$. By the second equation and considering $\alpha_1$,
$$r+s+2\alpha_1=0$$ so $$\alpha_1=\frac{r+s}{-2}$$
Similarly, considering $\alpha_2$,
$$r+s+2\alpha_2=0$$ so $$\alpha_2=\frac{r+s}{-2}$$
Hence $\alpha_1=\alpha_2$. To summarize, if a point is on two members of your family of sets, those two members are the same set. The contrapositive of this is: Two different members of your family of sets have no element in common. QED.