Let A(1, 1, 1), B(2, 3, 5), C(–1, 0, 2) be three points, then equation of a plane…
Mathematics · JEE Advanced · NTA Exams — Three Dimensional Geometry
Let A(1, 1, 1), B(2, 3, 5), C(–1, 0, 2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 from plane ABC
- \(2 x-3 y+z+2 \sqrt{14}=0\)
- \(2 x-3 y+z-\sqrt{14}=0\)
- 2x – 3y + z + 2 = 0
- 2x – 3y + z – 2 = 0
Answer
(A) 2 x-3 y+z+2 14 =0
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