JEE Main 2022 Mathematics question
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 :
Answer: (4)
Solution
The normal of the required plane is the cross product of the normals (2, 1, −5) and (3, 5, −7): (1·(−7) − (−5)(5), −5·3 − 2(−7), 2·5 − 1·3) = (18, −1, 7). The plane through (2, 3, −5) is 18(x −
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