If a, b, c are the x, y, z intercepts of a plane, and the perpendicular distance from origin is p, then:

If a, b, c are the x, y, z intercepts of a plane, and the perpendicular distance from origin is p, then:

• A. 1/a² + 1/b² + 1/c² = p²
• B. a² + b² + c² = p²
• C. 1/a² + 1/b² + 1/c² = 1/p²
• D. a + b + c = p

Answer: C) 1/a² + 1/b² + 1/c² = 1/p²

Explanation: Plane is x/a + y/b + z/c = 1. Distance p from origin is |−1| / √((1/a)² + (1/b)² + (1/c)²). Squaring gives p² = 1 / (1/a² + 1/b² + 1/c²), so 1/a² + 1/b² + 1/c² = 1/p².