If the centre of the sphere $x^{2} + y^{2} + z^{2} - 2 x - 4 y - 6 z = 0$
is (a,b,c) , find the value of a+b+c

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Solution

We are given the equation of a sphere and we want to find the centre of it. We know how to write the equation of sphere if we are given the centre and radius of the sphere. Let the centre be (a,b,c) and the radius be r. Then the equation of the sphere will be

$(x - a)^{2} + (y - b)^{2} + (z - c)^{2} = (r)^{2} \Rightarrow x^{2} + y^{2} + z^{2} - 2 a x - 2 b y - 2 c z + a^{2} + b^{2} + c^{2} = r^{2}$

Comparing this with the given equation $x^{2} + y^{2} + z^{2} - 2 x - 4 y - 6 z = 0$ ,

we get -2 = -2a, -4 = -2b and -6 = -2c $\Rightarrow$ a = 1, b = 2 and c = 3 So the centre of the sphere is (1,2,3) $\Rightarrow a + b + c = 1 + 2 + 3$ =6

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