Question 122

If $$a^2+b^2+c^2=2(a-2b-c-3)$$ then the value of a+b+c is

Given $$a^2+b^2+ c^2=2(a-2b-c-3)$$,

So, $$(a-1)^2+(b+2)^2+(c-1)^2=0$$

Hence, a=1, b=-2 and c=1

So, the sum of the equation is

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