Intersection of two planes, how to represent a line? If we have two planes: $$4x-y+3z-1=0$$  $$x-5y-z-2=0$$
and if we want to find a plane which contains the origin point and the intersection of the two planes given, how do we do it?
What my teacher did is form a system of two equations and subtract 4*II from I and he got:
$$19y+7z+7=0$$
What does that line represent? I understand it's a line in y-z plane, but is it really the intersection of these two planes? If it's not, why can't we find an intersection line just by equalizing the two planes? Why can't the line be represented in the three-dimensional space?
After that, the teacher found two points on the line:
$A (1, 0, -1)$ and $B (17, 7, -20)$
He found the first point by saying $y=0$ and solving, but what about the $B$ point, why didn't he simply say $z=0$?
After that, he made a cross product of vectors OA and OB and got the vector $n(7, 3, 7)$ which is a vector perpendicular to the unknown plane. Then he finds the plane as:
$$7(x-0)+3(y-0)+7(z-0) = 0$$
or
$$7x+3y+7z=0$$


What my teacher did is form a system of two equations and subtract
  4*II from I and he got: $$19y+7z+7=0$$

For the common points, both equations must hold, so we have a system
\begin{align}
4x-y+3z-1&=0 \\
x-5y-z-2&=0
\end{align}

What does that line represent? I understand it's a line in y-z plane,
  but is it really the intersection of these two planes? If it's not,
  why can't we find an intersection line just by equalizing the two
  planes? Why can't the line be represented in the three-dimensional
  space?

Just equating the function parts from the implicit equations $F_i(x,y,z) = 0$ like $F_1(x,y,z) = F_2(x,y,z)$ would describe a different solution set in general.
Your teacher eliminated one var in the first equation:
\begin{align}
19y+7z+7&=0 \\
x-5y-z-2&=0
\end{align}
Note that the other equation must be kept to have the same solution set.
The new first equation models yet another plane, but one that has the same intersection with the second equation, like the old first equation.



After that, the teacher found two points on the line: $A (1, 0, -1)$
  and $B (17, 7, -20)$ He found the first point by saying $y=0$ and
  solving, but what about the $B$ point, why didn't he simply say $z=0$?

I think it is guessing, the intersection might not have a solution with $y=0$. We can try it: 
$$
\begin{array}{rr}
x-5y-z-2 &= 0 \\
19y+7z+7 &= 0
\end{array}
\quad\to\quad
\begin{array}{rr}
x-z-2 &= 0 \\
7z+7 &= 0
\end{array}
$$
Here we get a solution if $z = -1$ and then $x = 1$. This is $A = (1,0,-1)$.
If we set $z=0$ then the resulting system turns into
$$
\begin{array}{rr}
x-5y-z-2 &= 0 \\
19y+7z+7 &= 0
\end{array}
\quad\to\quad
\begin{array}{rr}
x-5y-2 &= 0 \\
19y+7 &= 0
\end{array}
$$
Then we would first get $y = -7/19$ and $x = 2 + 5 (-7/19) = (38 - 35)/19 = 3/19$. This would be $P=(3/19,-7/19, 0)$.
The remaining part should be clear. The vector product of two linear independent vectors is orthogonal to both, and the origin and the intersection were part of the plane generated by $OA$ and $OB$.
