SSC CHSL 6 Dec 2015 Shift 1 Paper

Given : ${\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2=2\left(\mathit{a}-\mathit{b}-\mathit{c}\right)-3$
⇒ ${\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2=2\mathit{a}-2\mathit{b}-2\mathit{c}-3$
⇒ $\left({\mathit{a}}^2-2\mathit{a}\right)+\left({\mathit{b}}^2+2\mathit{b}\right)+\left({\mathit{c}}^2+2\mathit{c}\right)$ $=-\left(1+1+1\right)$
⇒ $\left({\mathit{a}}^2-2\mathit{a}+1\right)+\left({\mathit{b}}^2+2\mathit{b}+1\right)$ $+\left({\mathit{c}}^2+2\mathit{c}+1\right)=0$
⇒ ${\left(\mathit{a}-1\right)}^2+{\left(\mathit{b}+1\right)}^2$ $+{\left(\mathit{c}+1\right)}^2=0$
∵ 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
∴ (a + b + c) = 1 + (−1) + (−1) = −1