The problem
So recently in school, we should do a task somewhat like this (roughly translated):
Assign a system of linear equations to each drawing
Then, there were some systems of three linear equations (SLEs) where each equation was describing a plane in their coordinate form and some sketches of three planes in some relation (e.g. parallel or intersecting at 90°-angles.
My question
For some reason, I immediately knew that these planes:
belonged to this SLE:
$$ x_1 -3x_2 +2x_3 = -2 $$
$$ x_1 +3x_2 -2x_3 = 5 $$
$$-6x_2 + 4x_3 = 3$$
And it turned out to be true. In school, we proved this by determining the planes' intersecting lines and showing that they are parallel, but not identical.
However, I believe that it must be possible to show the planes are arranged like this without a lot of calculation. Since I immediately saw/"felt" that the planes described in the SLE must be arranged in the way they are in the picture (like a triangle). I could also determine the same "shape" on a similar question, so I do not believe that it was just coincidence.
What needs to be shown?
So we must show that the three planes described by the SLE cut each other in a way that I do not really know how to describe. They do not intersect with each other perpendicular (at least they don' have to to be arranged in a triangle), but there is no point in which all three planes intersect. If you were to put a line in the center of the triangle, it would be parallel to all planes.
The three planes do not share one intersecting line as it would be in this case:
(which was another drawing from the task, but is not relevant to this question except for that it has to be excluded)
My thoughts
If you were to look at the planes exactly from the direction in which the parallel line from the previous section leads, you would see something like this:
The red arrows represent the normal of each plane (they should be perpendicular). You can see that the normals somehow are part of one (new) plane. This is already given by the manner how the planes intersect with each other (as I described before).
If you now were to align your coordinate system in such a way that the plane in which the normals lie is the $x_1 x_2$-plane, each normals would have an $x_3$ value of $0$. If you were now to further align the coordinate axes so that the $x_1$-axis is identical to one of the normals (let's just choose the bottom one), the values of the normals would be somehow like this:
$n_1=\begin{pmatrix}
a \\
0 \\
0
\end{pmatrix}$ for the bottom normal
$n_2=\begin{pmatrix}
a \\
a \\
0
\end{pmatrix}$ for the upper right normal
and $n_3=\begin{pmatrix}
a \\
-a \\
0
\end{pmatrix}$ for the upper left normal
Of course, the planes do not have to be arranged in a way that the vectors line up so nicely that they are in one of the planes of our coordinate system.
However, in the SLE, I noticed the following:
-The three normals (we can simpla read the coefficients since the equations are in coordinate form) are $n_1=\begin{pmatrix}
1 \\
-3 \\
2
\end{pmatrix}$, $n_2=\begin{pmatrix}
1 \\
3 \\
-2
\end{pmatrix}$ and $n_3=\begin{pmatrix}
0 \\
-6 \\
4
\end{pmatrix}$.
As we can see, $n_1$ and $n_2$ have the same values for $x_1$ and that $x_2(n_1)=-x_2(n_2)$; $x_3(n_1)=-x_3(n_2)$
Also, $n_3$ is somewhat similar in that its $x_2$ and $x_3$ values are the same as the $x_2$ and $x_3$ values of $n_1$, but multiplied by the factor $2$.
I also noticed that $n_3$ has no $x_1$ value (or, more accurately, the value is $0$), while for $n_1$ and $n_2$, the value for $x_1$ is identical ($n_1=1$).
Conclusion
I feel like I am very close to a solution, I just don't know what to do with my thoughts/approaches regarding the normals of the planes.
Any help would be greatly appreciated.
How can I show that the three planes are arranged in this triangular-like shape by using their normals, i.e. without having to calculate the planes' intersection lines? (Probably we will need more than normals, but I believe that they are the starting point).
Update: I posted a new question that is related to this problem, but is (at least in my opinion) not the same question.
Best Answer
If you write your systems of equations as a matrix as follows: $$A \vec{x} = \begin{bmatrix} 1 & -3 & 2 \\ 1 & 3 & -2 \\ 0 & -6 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2 \\ 5 \\ 3\end{bmatrix} = \vec{b}$$ then here is a (perhaps) quicker way to determine if the picture looks like the triangle. Note: I don't know how comfortable you are with basic linear algebra concepts, but you only need them to understand the proof of why this is correct. You can apply the method without any understanding of them.
Example: I'll work thru the steps above with the example you've written.
Steps $1$ and $2$. I can immediately notice that none of normal vectors of the planes are parallel. So we proceed to step $3$.
Step $3$. We can calculate $$\det(A) = (1)(12 - 12) - (-3)(4 - 0) + 2(-6 - 0) = 0$$ so we proceed to step $4$. Note that if you were able to observe that the third row of $A$ was a linear combination of the first and second row (the third row is simply the first row minus the second row) or that the third column was a multiple of the second column, you could immediately skip to step $4$.
Step $4$. We can notice that none of the columns are zeroes (case a), but in fact the last two columns are multiples of each other. So case b) applies here, and we have to exchange one of the last two columns with $\vec{b}$ for the process to be correct. Let's replace the last column of $A$ with $\vec{b}$ to obtain $A'$: $$A' = \begin{bmatrix} 1 & -3 & -2 \\ 1 & 3 & 5 \\ 0 & -6 & 3 \end{bmatrix}$$ and we can calculate $$\det (A') = (1)(9 + 30) - (-3)(3 - 0) + (-2)(-6 - 0) = 29 + 9 + 12 = 60 \neq 0$$ and hence we can conclude we have the "triangle" configuration.
Conclusion: I think this method is somewhat easier than calculating the three intersection lines. It requires you to calculate two determinants of $3 \times 3$ matrices instead.