<p>The number of planes that are equidistant from four non-coplanar points is</p>
Step-by-Step Solution
Key Concept: Four non-coplanar points form a tetrahedron. A plane equidistant from all four vertices must be positioned such that all four points are equidistant from it. This occurs at planes that are perpendicular bisectors of segments connecting pairs of points, and crucially, at planes that bisect the tetrahedron symmetrically.
Step 1: For a plane to be equidistant from 4 non-coplanar points, each point must lie at the same perpendicular distance from the plane. Step 2: Consider a tetrahedron ABCD. The planes equidistant from all 4 vertices are: <ul><li>The 6 planes that are perpendicular bisectors of the 6 edges (each passes through the midpoint of an edge and is perpendicular to it). However, these planes are equidistant from the two endpoints of each edge, not all 4 vertices.</li><li>The key insight: We need planes where d(A,π) = d(B,π) = d(C,π) = d(D,π).</li></ul> Step 3: For any tetrahedron, there are exactly 7 planes equidistant from all four vertices: <ul><li> 4 planes: Each passes through the midpoint of one edge and the midpoints of the two opposite edges (these are planes of symmetry for certain configurations)</li><li> 3 planes: Each passes through two opposite edges' midpoints (perpendicular bisector planes of opposite edge pairs)</li></ul> Step 4: Through systematic analysis using coordinate geometry or symmetry arguments, the total count is 7 planes . ∴ Answer: 7
Correct Answer: 7