The equation of the sphere one of whose diameter has end points $(1, 2, 4)$ and $(3, 0, 2)$ is

x^{2} + y^{2} + z^{2} + 4 x + 6 y + 8 z + 11 = 0

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x^{2} + y^{2} + z^{2} + 4 x + 4 y + 8 z - 11 = 0

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x^{2} + y^{2} + z^{2} - 4 x - 2 y - 6 z + 11 = 0

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x^{2} + y^{2} + z^{2} - 4 x - 2 y + 6 z - 11 = 0

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Solution

The correct option is C $x^{2} + y^{2} + z^{2} - 4 x - 2 y - 6 z + 11 = 0$
End point of diameter of sphere are $(1, 2, 4)$ and $(3, 0, 2)$ .
So centre of sphere is $(2, 1, 3)$ and radius is $= \frac{1}{2} \sqrt{(1 - 3)^{2} + (2 - 0)^{2} + (4 - 2)^{2}} = \sqrt{3}$
Hence, equation of required sphere is
$(x - 2)^{2} + (y - 1)^{2} + (z - 3)^{2} = 3$
$\Rightarrow x^{2} + y^{2} + z^{2} - 4 x - 2 y - 6 y + 11 = 0$

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