When solving any system of equations (2 unknown and 2 equations, 3 unknowns and three equations, 4 unknowns and 4 equations, and so on...) it's entirely possible you'll wind up with a contradiction, like 0=8, or some other nonsensical result. In this case, all it means that there is no solution that will satisfy all the equations. So, your assertion is correct.
That means that there is no single point that is on all three plans. As you also state, it could be that one plane intersects the other two, which may well be parallel. The easiest way to check that is to take the dot product of the vectors represented by two of the planes. For instance, say the first two planes are given by the equations:
A1x + B1y + C1z = D1
A2x + B2y + C2z = D2
So we could evaluate the dot product of the vector <A1 B1 C1> with <A2 B2 C2>:
<A1 B1 C1> ∙ <A2 B2 C2>
which evaluates to
(A1 * A2) + (B1 * B2) + (C1 * C2)
which you can obtain a numerical solution for.
The two planes are perpendicular to one another if this evaluates to zero. Otherwise, they are not perpendicular to one another.
The two planes are parallel if A1,B1, and C1 are all the same multiple of A2, B2, and C2 respectively. For example, the two planes
6x - 3y + (3/2)z = 0
12x - 6y + 3z = 0
Are parallel, because the second plane's coefficients are all factors of 2 of the first.
You can evaluate all the planes in this manner to see which two (if any) are parallel, perpendicular, or otherwise to one another.