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

11 \frac{4}{13}

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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 $13$
Centre and radius of the given sphere are $(- 2, 1, 3)$ and $\sqrt{2^{2} + 1^{2} + 3^{2} + 155} = 13$ respectively.

Now, the distance between centre of sphere to the given plane is given by,
$d = ∣ ∣ ∣ ∣ \frac{12 (- 2) + 4 (1) + 3 (3) - 327}{\sqrt{12^{2} + 4^{2} + 3^{2}}} ∣ ∣ ∣ ∣ = \frac{338}{13} = 26$
Hence, shortest distance from sphere to the plane is $= 26 - 13 = 13$ .

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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

π : x + 2 y + 3 z = p

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

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p = 6

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p = 26

x^{2} + y^{2} + z^{2} + 7 x - 2 y - z = 13

x^{2} + y^{2} + z^{2} - 3 x + 3 y + 4_{Z} = 8

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