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

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11 \frac{4}{13}

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Solution

The correct option is C 13
$∵$ Shortest distance = Perpendicular distance from centre - r
Now, perpendicular distance
$= ∣ ∣ ∣ \frac{- 2 \times 12 + 4 \times 1 + 3 \times 3 - 327}{\sqrt{144 + 9 + 16}} ∣ ∣ ∣ = 26$
$∴$ Shortest distance $= 26 - \sqrt{4 + 1 + 9 + 155}, [∵ 26 - r]$
=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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