Sphere $3 x^{2} + 3 y^{2} + 3 z^{2} - 6 x - 12 y + 6 z + 2 = 0$ has centre ________.

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Solution

Equation of the sphere is given as:
$3 x^{2} + 3 y^{2} + 3 z^{2} - 6 x - 12 y + 6 z + 2 = 0$

Converting this equation to the standard form, we get
$3 ((x - 1)^{2} - 1 + (y - 2)^{2} - 4 + (z + 1)^{2} - 1) + 2 = 0$
$\Rightarrow (x - 1)^{2} + (y - 2)^{2} + (z + 1)^{2} = 16$

Therefore, Centre is $(1, 2, - 1)$ and radius is $4$ units.

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