For definiteness, I'll assume you're asking about planes in Euclidean space, either R3, or Rn with n≥4.

The intersection of two planes in R3 can be:

- Empty (if the planes are parallel and distinct);
- A line (the "generic" case of non-parallel planes); or
- A plane (if the planes coincide).

The tools needed for a proof are normally developed in a first linear

algebra course. The key points are that non-parallel planes in R3 intersect; the intersection is an "affine subspace" (a translate of a vector subspace); and if k≤2 denotes the dimension of a non-empty intersection, then the planes span an affine subspace of dimension 4−k≤3=dim(R3). That's why the intersection of two planes in R3 cannot be a point (k=0).

Any of the preceding can happen in Rn with n≥4, since R3 be be embedded as an affine subspace. But now there are additional possibilities:

- The planes
P1={(x1,x2,0,0):x1,x2 real},P2={(0,0,x3,x4):x3,x4 real}

intersect at the origin, and nowhere else. - The planes P1 and
P3={(0,x2,1,x4):x2,x4 real}

are not parallel (in the sense that neither is a translate of the other), but they do not intersect.

The planes P1 and P3 are "partially parallel" in the sense that there exist parallel lines ℓ1⊂P1 and ℓ3⊂P3. This turns out to be true for every pair of disjoint planes in R4.

In R5, there exist "totally skew" planes, such as

P4={(x1,x2,0,0,0)},P5=(0,0,1,x4,x5)}