Minimum number of non-coplanar forces required to keep an object in equilibrium 
The minimum number of non-coplanar forces that can keep a particle in equilibrium is:
(a) 1
(b) 2
(c) 3
(d) 4

Answer given is option $(d)$ , i.e $4$. But can’t it be $(c)$ , i.e $3$ too?
Suppose I have three forces:
$F_1= 3i +4j$
$F_2= -3i +5k$
$F_3= -4j -5k$
$F_1$ lies in x-y plane, $F_2$ in x-z plane and $F_3$ in z-y plane. So all of them are non coplanar. But sum of all of them results in $0$. So shouldn’t three forces be enough to keep an object in equilibrium?
 A: Actually, the forces $\vec{F_1}, \vec{F_2}, \vec{F_3}$ defined by the points $(3,4,0)$ , $(-3,0,5)$ and $(0,-4,-5)$ are coplanar.
Even though $\vec{F_1}, \vec{F_2}, \vec{F_3}$ algebraically sum to $\vec 0$ they are nevertheless coplanar.
In fact, if you do a little algebra (solve three simultaneous equations) or calculate the cross product $(\vec{F_2}-\vec{F_1})\times (\vec{F_3}-\vec{F_1})$ (which gives you the vector normal to this plane) you deduce that $(3,4,0)$, $(-3,0,5)$ and $(0,-4,-5)$ lie on the plane defined by the equation $$20x-15y+12z=0$$
So the rule still holds and you would actually need $3+1$ non coplanar forces for equilibrium, where the forth force would need to be equal and opposite to the sum of the first three  forces.
Note that three coplanar forces, or even two coplanar forces can sum to zero.
A: The question seems to be worded incorrectly.

*

*If there is a known applied force, then 3  force magnitudes are linearly independent directions are needed to oppose the force. I can see that 3+1 would be the answer if you include the applied force.


*Or one single force in opposite direction and magnitude if the directions are arbitrary. In this case, the answer is 1+1.
The issue that I have is that it is nowhere stated that there is one force acting on the particle. There could be 12 forces acting, and needed 3 reaction directions to keep stable for an answer of 12+3 = 15. The situation is very ambiguous as stated.
