The equation of the sphere whose centre is the point $(2, 3, 1)$ and which touch the plane $x + y + z = 0$

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

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

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

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

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Solution

The correct option is B $x^{2} + y^{2} + z^{2} - 4 x - 6 y - 2 z + 2 = 0$
Radius of the sphere is $r = ∣ ∣ ∣ \frac{1 + 2 + 3}{1^{2} + 1^{2} + 1^{2}} ∣ ∣ ∣ = 2 \sqrt{3}$
Therefore the equation of sphere with centre $(2, 3, 1)$ is given by,
$(x - 2)^{2} + (y - 3)^{2} + (z - 1)^{2} = 12$
$\Rightarrow x^{2} + y^{2} + z^{2} - 4 x - 6 y - 2 y + 2 = 0$

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