How to find the union&intersection of two lines by their equations? I will try to be as clear as possible concerning my confusion, and I will use some examples(several ones).
Case number 1.
Assume two equations(in cartesian form) of two planes.
$2x+2y-5z+2=0$ and $x-y+z=0$
Now,we need to find their vectors.
For the first on, we get: {1,-1,0}, {0,5/2,1} and {1,0,2/5}.
For the second equation, we get: {1,1,0}, {0,1,1},and {1,0,-1}.
Now, I have a hard time understanding how I have to figure out where is their intersection and their union?
Case 2:
Assume the one of the previous planes $x-y+z=0$ and the line  $x-y=0$. How do I find the intersection and union of these two?

First of all: $Ax+By+Cz+D=0$ is plane equation.
Case 1:
Intersection of to planes is line. To find equation of that line you have to solve system of equations:
$$
2x+2y-5z+2=0\\
x-y+z=0 \Rightarrow x=y-z \\
$$
If we substitute second equation into first we got
$$
2(y-z)+2y-5z+2=0 \Rightarrow 4y-7z+2=0 \Rightarrow y=\frac{7z-2}{4}
$$
and for x
$$
x=\frac{7z-2}{4}-z=\frac{3z-2}{4}
$$
Now we have parametric equation of the intersection
$$
x=-\frac{1}{2}+\frac{3}{4}z\\
y=-\frac{1}{2}+\frac{7}{4}z\\
z=0+1z
$$
It's line equation which can be written in another form
$$
\frac{x+\frac{1}{2}}{\frac{3}{4}}=\frac{y+\frac{1}{2}}{\frac{7}{4}}=\frac{z}{1}
$$
If we multiply denominators by 4 we get shorter equation
$$
\frac{x+\frac{1}{2}}{3}=\frac{y+\frac{1}{2}}{7}=\frac{z}{4}
$$
Case 2:
$x-y=0$ is not line equation - it's plane equation $x-y+0z=0$. So this case is also intersection of two planes - like case 1.
Solution:
$$
x-y+z=0 \Rightarrow x=y-z\\
x-y=0 \Rightarrow y-z-y=0 \Rightarrow z=0
$$
Using $x=y-z$ we have $x=y$. So, intersection is a line whose equation is
$$
x=0+1y\\
y=0+1y\\
z=0+0y
$$
or in another form
$$
\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{0}
$$
