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

11 \frac{3}{4}

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13

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39

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26

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Solution

The correct option is B $13$
The centre of the sphere is (-2, 1, 3) and its radius is $\sqrt{4 + 1 + 9 + 155} = 13$ .
Length of the perpendicular from the centre of the sphere on the plane is $∣ ∣ ∣ \frac{- 24 + 4 + 9 - 327}{\sqrt{144 + 16 + 9}} ∣ ∣ ∣ = \frac{338}{13} = 26$
So the plane is outside the sphere and the required distance is equal to 26 - 13 = 13.

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Similar questions

12 x + 64 y + 3 z = 327

x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155

x^{2} + y^{2} + z^{2} = 64

x^{2} + y^{2} + z^{2} - 12 x + 4 y - 6 z + 48 = 0

x^{2} + y^{2} + z^{2} = 64

x^{2} + y^{2} + z^{2} - 12 x + 4 y - 6 z + 48 = 0

x^{2} + y^{2} + z^{2} - 2 x + 4 y - 6 z + 7 = 0

x^{2} + y^{2} + z^{2} = 5, x + 2 y + 3 z = 3

4 x + 3 y -

15 = 0

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