Why is it ok to choose an arbitrary point when calculating the intersection of two planes? When I go through examples of calculating the intersection of two planes, there seems to be a convention of choosing an arbitrary point in order to solve the linear equations in question and get a particular point on the intersection line.
Let's take an example from a previously-asked question, where the given equations were:

$x + 2y + z - 1 = 0$
$2x + 3y - 2z + 2 = 0$

And the relevant part in the second-most voted answer goes as follows:

Next, we need to find a particular point on the line. We can try $y=0$
  and solve the resulting system of linear
  equations:$$\begin{align}x+z-1&=&0\\2x-2z+2&=&0\end{align}$$ giving
  $x=0, z=1$

My question: 
How does one know what is the correct point to choose, and how does one validates that the chosen point is correct?
Also, if the chosen point is wrong, how does one successfully guess the next point? 
I've found another answer that seems to be very relevant, but I can't explain it:

Sometimes the line of intersection happens to be parallel to the z=0
  plane. In that case you could try y=0 or x=0. (One of these is sure to
  work.)

I'm a beginner, so that's a fundamental thing, probably trivial, I don't get: Why is that true?
 A: The intersection of two non-parallel planes is a line. That line can be given by the equation $\mathbf{r}(t)=\mathbf{r}_0 + t\mathbf{L}$ where $\mathbf{L}$ is a direction vector. If $\mathbf{L}$ has all nonzero components (the "typical" situation), then any given one of the variables will be zero at some point on the line (in general different points for each). You can pick the $\mathbf{r}_0$ of the line to be one of those points. This is just a matter of convenience, it doesn't have some deep significance.
On the other hand, if the direction vector has a zero component (which geometrically means that the line is directed parallel to one of the coordinate planes) and you pick that coordinate to set equal to zero then you will probably get an inconsistent system. You can just pick a different coordinate to set equal to zero; one of the three will necessarily work.
Alternatively, you can approach solving the two linear equations in three unknowns directly by using Gaussian elimination (which you have probably learned at some point for systems of two linear equations in two unknowns). In this problem you can eliminate $x$ in the second equation, changing it to $-y-4z+4=0$, and then use that equation to eliminate $y$ in the first equation, changing it to $x-7z+7=0$. Now you read these new equations as $x=7z-7,y=-4z+4$ where $z$ is free, which can be easily translated into an equation of the line. If the line turns out to be parallel to a coordinate plane, this procedure will tell you that directly without needing to guess-and-check.
A: There is not "the" correct point. The goal is to find any specific point on the line.
The solver has chosen to set $y=0$ because that results in two equations in two unknowns ($x$ and $z$), which is an easy system to solve.
You could just as easily have picked any other value for any of the variables, but the numbers might have been not as easy to work with (that variable would become a constant, and you would have to move it to the other side in each equation, etc.)--so it's easier if you just make a variable vanish by setting it equal to zero.
The reason specifying a particular value (like $0$) for one of the variables usually works is that the "typical" line is not parallel to any of the axes, so it will contain points with any given value (the one you specify) for any one of the coordinates. But if you are unlucky and the line line is parallel to one of the axes, then there may be no point on the line with that particular coordinate value. In that case, you would find that the resulting system of 2 equations in 2 unknowns has no solution; then you would just do it over, picking a specific value for one of the other variables instead.
A: The intersection line can be parametrized as
$$g: \mathbb{R} \longrightarrow \mathbb{R}^3,\ t \longmapsto a + tv,$$
where $a$ is an arbitrary point lying on the line and $v$ is a directional vector. In the answer you quoted, $v$ was already calculated, so only $a$ misses. We try $y = 0$ and get the solution $a = (x, y, z) = (0, 0, 1)$. In general, you always need one particular solution of the corresponding linear equation system. 
Sometimes guessing a variable does not immediately lead you to success, as it can be seen for instance in the second answer you quoted with $z = 0$. If this is the case, you can try to set another variable to $0$. This will work (in the case $z = 0$, otherwise similar), since every line in the $(x, y)$-plane goes through at least one point with coordinates $(x, 0)$ or $(0, y)$ (draw the picture).
