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CGP EDU Academic Team
Published on: August 14, 2026
Let the plane ax + by + cz = d pass through (2, 3, -5) and is perpendicular to the planes 2x + y - 5z = 10 and 3x + 5y - 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to
Text Solution
Verified by ExpertsThe correct answer is:
D
DR'S normal of plane
equation of plane
18x-y + 7z = d
It passes through (2, 3, -5)
36-3-35 = d 
d =-2
Eq n of plane
18x-y + 7z = -2
-18x + y - 7z = 2
a=-18,b=1,c = -7,d=2
a + 7b + c + 20d = -18 + 7-7 +40 = 22
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