The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x^2 + y^2 + z^2 + 4x - 2y - 6z = 155 is: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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The shortest distance from the plane 12 x + 4 y + 3 z = 327 to the sphere x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155 is:

Solution & Step-by-step Explanation

Sphere: x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z - 155 = 0 . Center C = (- 2, 1, 3) . Radius R = (- 2)^{2} + 1^{2} + 3^{2} - (- 155) = 4 + 1 + 9 + 155 = 169 = 13 . Distance from C (- 2, 1, 3) to plane 12 x + 4 y + 3 z - 327 = 0 : p = \frac{∣12 ( - 2 ) + 4 ( 1 ) + 3 ( 3 ) - 327∣}{1 2 ^{2} + 4 ^{2} + 3 ^{2}} = \frac{∣ - 24 + 4 + 9 - 327∣}{13} = \frac{∣ - 338∣}{13} = 26 Shortest distance from plane to sphere surface = p - R = 26 - 13 = 13 .

Try it yourself before checking the explanation above.

The shortest distance from the plane 12 x + 4 y + 3 z = 327 to the sphere x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155 is:

26

11 \frac{4}{13}

13

39

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