A plane intersects a regular tetrahedron in such a way that the intersection forms an equilateral triangle. How many such planes can do this?

The logic follows here is:

Here, we need to identify how many planes can intersect a regular tetrahedron in such a way that the intersection forms an equilateral triangle. A regular tetrahedron has four faces, each of which is an equilateral triangle. When a plane intersects the tetrahedron to form an equilateral triangle, it must be parallel to one of these faces and pass through the midpoints of the opposite edges.

1. Consider the first face of the tetrahedron. A plane parallel to this face and passing through the midpoints of the opposite edges will intersect the tetrahedron in an equilateral triangle.

2. Similarly, for the second, third, and fourth faces of the tetrahedron, we can find such planes. Since a regular tetrahedron has four faces, there are exactly four such planes that can intersect the tetrahedron to form an equilateral triangle.

Hence, the correct answer is "Option number 2".