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
$$7x+3y+7z=0$$
 A: 
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 $OB$.
A: Your teacher simply did row reduction in order to eliminate one unknown. $19y+7z+7=0$ is simply the equation of a plane that passes through the itersection line of the two given planes. 
Remember  a line requires two linear equations in $3$-space.
For the point $B$, I guess he set $y=7$ ih the equation so as to prevent obtaining denominators for $z$.
Added: Let's denote $n_1=\begin{bmatrix}4\\-1\\3\end{bmatrix}$ and  $n_2=\begin{bmatrix}1\\-5\\-1\end{bmatrix}$ be the normal vectors to the planes. Then $n_1\times n_2=\begin{bmatrix}16\\7\\-19\end{bmatrix}$ is a directing vector of the intersection line, so that its parametric equations are:
$$\begin{cases}
x=1+16t\\ y=7t\\z=-1-19t
\end{cases}$$
A: There is more easy way to find the required plane, 
Let $S_1=0, S_2=0$ be given two plane, then equation of plane containing these two plane is given by, $$S_1+\alpha S_2=0$$. By some more given condition we can find the value of $\alpha$, then by putting value of $\alpha$ in above eqution we will get required plane. 
Now in your case, $$\begin{align}
4x-y+3z-1&+\alpha (x-5y-z-2)&=0
\end{align}$$
this plane passing through the origin, we have
$$\alpha=\frac{-1}{2}$$
Putting the value of $\alpha$ in the equation and simplifying  we get the required result as, $$7x+3y+7z=0$$
A: What he did: He decided to find the point lying on the intersection of the planes whose $x$ coordinate is $1$.  So he had to solve the system $$ 3z-y=-3$$ $$5y+z=-1$$. The solution is $y=0$, $z=-1$ so $(1,0,-1)$ is a point on the intersection. Then, he plugged in $x=17$ and he got the point ($17,7,-20$). The choice of $1$ and $17$ was random, as you could fix any two values of $x$. Having found those two points, he found three non colinear points contained in the plane ($O,A$ and $B$). Then he said that for any point $M(x,y,z)$ in the plane $\vec{n} \cdot \vec{OM} =0$ where $\vec{n}$ is normal to the plane. So he found a normal vector, mainly $\vec{OA} \times \vec{OB}$.
