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The distance of the point (2, 1, -1) from the plane 2x - y + 2z + 5 = 0 is

A6/√3
B6/2
C6/3
D6
Answer & Solution
Correct answer: C. 6/3
1. Use d = |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²). 2. Numerator: |2(2) - 1(1) + 2(-1) + 5| = |4 - 1 - 2 + 5| = 6. 3. Denominator: √(4 + 1 + 4) = √9 = 3. 4. d = 6/3 = 2. _Source: syllabus Maths Part 2 Ch 11 "Three Dimensional Geometry", §11.9_
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