3 eagerbugs can be on these planes:
!!z+2y-3x=0, z+2x+y=0, z-2x+8y=0!!
Can they meet at a point other than origin?
Looking at the 3 equations we can say that !!(0,0,0)!! is definitely a solution of the 3 equations.
We just need to find if they intersect at one more point.
There are more than 1 ways to find it.
Simplest approach would be to solve the 3 equations. So on subtracting second from first we get !!y=5x!! and thrid from first we get !!-6y-x=0!!, this can happen only if !!y=0,x=0!!, so !!z=0!!. Hence there is only 1 solution.
Second approach is based on the cramers rule which is used to solve a system of linear equations. We get
So this equation has only one trivial solution.