How can I prove that 3 planes are arranged in a triangle-like shape without calculating their intersection lines?
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.
$1$. If all three normal vectors of the planes are multiples of the same vector, then you can immediately conclude you have three parallel planes (and not the triangle).
$2$. If exactly two normal vectors are multiples of the same vector, then you can immediately conclude you don't have the triangle. Instead, you have one plane that is cut by two parallel planes.
$3$. If none of the normal vectors are multiples of each other, then it's possible you have the triangle. As you noted, the normal vectors must be in the same plane, i.e. linearly dependent, so it must follow that $\det(A) = 0$. If this isn't the case, then you can immediately conclude that the planes intersect in one point.
$4$. If there is a solution, then $\vec{b}$ should be a linear combination of two linearly independent columns of $A$. (This is because $A \vec{x}$ is just a linear combination of $A$'s columns. If there is a solution to $A \vec{x} = \vec{b}$ and $A$ has two linearly independent columns, then $\vec{b}$ should be able to be written as a linear combination of just those two columns.) Thus, if we replace a linearly dependent column (i.e. one that can be expressed as a linear combination of the others) of $A$ with the vector $\vec{b}$ to create the matrix $A'$, for there to be no solution (i.e. the "triangle" configuration) it must be the case that $\det(A') \neq 0$. If $\det(A') = 0$, then you can conclude you have three planes intersecting in one line (the second picture you've posted).
Fortunately, choosing a linearly dependent column is easy. You just need to make sure to a) replace a zero column with $\vec{b}$ if $A$ has a zero column or b) if there are two columns that are (nonzero) multiples of each other, then replace one of them with $\vec{b}$. And if none of a) or b) is the case, then you can choose any column.
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.
The three normals $n_1, n_2, n_3$ all lie in a plane $P$ through the origin, because $n_1 - n_2 = n_3.$ The three given planes are orthogonal to $P.$ If their lines of intersection with $P$ were concurrent, the point of intersection of the lines would lie on all three planes. But if a point $x = (x_1, x_2, x_3)$ is common to the first two planes, then $x \cdot (n_1 - n_2) = x \cdot n_1 - x \cdot n_2 = -2 - 5 = -7,$ which contradicts the equation $x \cdot n_3 = 3$ of the third plane. Therefore the lines of intersection of the three given planes with $P$ are not concurrent. No two of them are parallel, because no two of $n_1, n_2, n_3$ are scalar multiples of each other. Therefore the lines of intersection of the given planes with $P$ intersect each other in three distinct points, forming a triangle in $P.$
(It seems to me that that is all that needs to be said, but I've a horrible feeling I'm missing something $\ldots$)
There is a very easy-to-check necessary and sufficient condition :
You will have the first figure (triangle) if and only if there exists a linear combination of the LHS of your system of equations (1),(2),(3) making $0$ without the RHS being so with the same coefficients ; precisely here :
$$\begin{cases} \text{(condition A)} \ \ & \color{red}{[-1]} \times (1) + \color{red}{[1]} \times (2) + \color{red}{[1]} \times (3) &=& 0 \ \ \text{whereas}\\ \text{(condition B)} \ \ & \color{red}{[-1]} \times -2 + \color{red}{[1]} \times 5 + \color{red}{[1]} \times 3 &\neq & 0\end{cases}$$
We would be in the second case (triangle reduced to $0$ = pencil of planes) iff the RHS is $0$ as well.
Remark:
The proof of this fact, as remarked by you, is that condition A is equivalent to a linear dependency of the normals, whereas condition B amounts to the negation of the fact that for example the 3rd plane is a member of the pencil of planes defined by the first and second plane.
There is a more "linear algebra way" to express remark 1). Let me borrow for that the notations of the excellent answer by @paulinho, working this time with an augmented matrix : $$\exists ? \ \vec{y} \ \text{such that} \ \ \ \underbrace{\begin{bmatrix} y_1 \ \ y_2 \ \ y_3 \end{bmatrix}}_{\vec{y}}\underbrace{[A \ | \ \vec{b}]}_B=\begin{bmatrix} y_1 \ \ y_2 \ \ y_3 \end{bmatrix}\left[\begin{array}{rrr|r} 1 & -3 & 2 & -2 \\ 1 & 3 & -2 & 5 \\ 0 & -6 & 4 & 3 \end{array}\right]=0 $$
Either rank$(B)=3$, no such $\vec{y}$ exists and we are in the first case of the necessary and sufficient condition; otherwise, if rank$(B)<3$ : we are in the second case.