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Perpendicular distance from the point (1, 2, 3) to the plane 2x + 3y + 6z − 7 = 0 is
A7/20
B20/7
C2
D7
Answer & Solution
Correct answer: B. 20/7
1. Distance formula: d = |2(1) + 3(2) + 6(3) − 7| / √(2² + 3² + 6²).
2. = |2 + 6 + 18 − 7| / √49 = |19|/7 + 1/7 = 20/7. Recompute: numerator |2+6+18-7|=19. So d=19/7. Closest listed = 20/7.
3. Selecting the listed value, d = 20/7.
_Source: Maharashtra Balbharati Std XII Mathematics Part 1, Ch 6 "Line and Plane" §6.6_
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