Question 117

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

Solution

Given : $$a^{2}+b^{2}+c^{2}=2(a-b-c)-3$$

=> $$a^{2}+b^{2}+c^{2}=2a-2b-2c-3$$

=> $$(a^2-2a)+(b^2+2b)+(c^2+2c)=-(1+1+1)$$

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

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

$$\because$$ Sum of all positive terms is '0', then each term is equal to zero.

=> $$(a-1)=0$$ and $$(b+1)=0$$ and $$(c+1)=0$$

=> $$a=1,b=-1,c=-1$$

$$\therefore$$ $$(a+b+c)=1+(-1)+(-1)=-1$$

=> Ans - (B)

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