Conclusion that can be drawn from 4 points in space whose angles are $90^{\circ}$ Four points A, B, C and D are in space such that angles $A\hat BC, B\hat CD, C\hat DA$ and $D\hat AB$ are all right angles, then


*

*A, B, C, D cannot be coplanar

*A, B, C, D are necessarily coplanar.

*A, B, C, D may or may not be coplanar.

*No such points exist.


I got an answer by pure imagination. Please provide a proof/reasoning so as to why you feel your answer is correct.
 A: From $\angle{ABC}=90^\circ$, we may suppose that 
$$A(0,a,0),\quad B(0,0,0),\quad C(c,0,0)$$
It follows from $\angle{BCD}=90^\circ$ that the $x$-coordinate of $D$ is $c$.
It follows from $\angle{DAB}=90^\circ$ that the $y$-coordinate of $D$ is $a$.
Let $D(c,a,d)$. By the law of cosines, 
$$AC^2=DA^2+DC^2-2\cdot DA\cdot DC\cos\angle{CDA},$$
$$\Rightarrow \quad c^2+a^2=c^2+d^2+a^2+d^2\quad\Rightarrow\quad d=0.$$
The four points are on the plane $z=0$.
A: Only the first three right angles are adequate to establish that the 4 points belong to a common (developable) plane. However as an aside, the four points need not strictly be confined to a plane in general.
We can  apply  Gauß - Bonnet theorem when going around a loop along geodesic arcs in 3 space. If the sum of angles turned at each intersection is $ \Sigma \varphi $ , integral curvature is $ \int K d A $, we have 
$$ \Sigma  \varphi +  \int K d A = 2 \pi +  \int K d A = 2 \pi , K =0, $$
we should have not only a plane but cones/cylinders and three types of developables  associated with curves on surfaces.
A: As ABC is right angle, line segment 
1) line segment AB & BC must be on same plane.
2) line segment BC & CD must be on same plane
3) line segment CD & DA must be on same plane
4) line segment DA & AB must be on same plane.
(any two lines are always on same plane)
From points 1 and 2, AB,BC and DA has to share a common plane. From either point 2 or 3, CD must also be in same plane.
So all the line segments must be on same plane and so all points are coplanar.
