I think question is "whether they can form a right angled triangle" without rotation, so $2$ conditions only are to check:
- vectors can be placed "in one plane": one of $\vec{A}\pm \vec{B}\pm \vec{C}$ must be $0$;
- $2$ of them form $90^\circ$: dot product of some $2$ vectors is $0$.
If these $2$ conditions are true, then Pythagoras theorem is excessive checking.
Here we have:
$\vec{A}-\vec{B}-\vec{C}=0$,
$\vec{A}\cdot\vec{C}=0$.
That say that vectors $\vec{A},\vec{B},\vec{C}$ form right triangle.
Example without $1$st condition:
$\vec{A}=(3,0,0),\quad\vec{B}=(0,4,0),\quad\vec{C}=(0,0,5)$. (don't lie in one plane).
Example without $2$nd condition:
$\vec{A}=(3,3,3),\quad\vec{B}=(1,2,1),\quad\vec{C}=(2,1,2)$. (there are no angle $90^\circ$).
A general rule for finding the counterclockwise angle from the positive $x$ axis to the direction of a two-dimensional vector is explained in
How to convert components into an angle directly (for vectors)?
The basic idea is that you first take the arc tangent of the $y$-component
divided by the $x$-component:
$$
\theta_1 = \arctan\frac{a_y}{a_x}.
$$
One complication is that this procedure gives the same resulting $\theta_1$
for the vector $\langle a_x,a_y \rangle$ and the vector $\langle -a_x,-a_y \rangle,$ which are two vectors in directions $180$ degrees apart.
So $\theta_1$ may be in the direction you want, or in the direction
$180$ degrees opposite.
In fact $\theta_1$ will be in the correct direction whenever $a_x > 0,$
since that is how the arc tangent function is designed to work.
(It always produces angles in the range $-\frac\pi2$ to $\frac\pi2$ radians,
that is, $-90$ to $90$ degrees,
which correspond to vectors with positive $x$-components.)
The cases where $\theta_1$ is wrong are precisely the cases where
$a_x < 0,$ so if $a_x < 0$ (as it is in your particular question)
you have to add $180$ degrees to $\theta_1.$
A second complication (which does not actually come up in your particular
question) occurs because the procedures given above produce a result in the range $-90$ degrees to $270$ degrees, but people usually want an answer in the range $-180$ to $180$ or $0$ to $360.$ The solution to this, of course, is to add or subtract $360$ degrees as needed to get an answer in the desired angle range.
Best Answer
If the vectors form any closed polygon then traversing them all should bring you back where you started:
$$ \vec{a} + \vec{b} + \vec{c} = 0 $$
In this case the vectors do not form a triangle, though they do if you replace $\vec{b}$ by $-\vec{b}$ i.e. make $\vec{b}$ point in the other direction.