The equation of the sphere with centre at $(- 1, 2, 3)$ and which passes through $(1, - 1, 2)$

x^{2} + y^{2} + z^{2} + 2 x - 4 y - 6 z - 14 = 0

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x^{2} + y^{2} + z^{2} - 2 x + 4 y + 6 z - 14 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

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

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Solution

The correct option is D $x^{2} + y^{2} + z^{2} + 2 x - 4 y - 6 z = 0$
Given centre of the sphere is $(- 1, 2, 3)$ and it also passes through $(1, - 1, 2)$
Therefore radius of the sphere is $= \sqrt{(1 + 1)^{2} + (- 1 - 2)^{2} + (2 - 3)^{2}} = \sqrt{14}$
Hence, equation of the required sphere is,
$(x + 1)^{2} + (y - 2)^{2} + (z - 3)^{2} = 14$
$\Rightarrow x^{2} + y^{2} + z^{2} + 2 x - 4 y - 6 z = 0$
Hence, option D is correct.

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