SSC CGL 19 Oct 2014 Shift 2 Paper

3 (${\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2$) = ${\left(\mathit{a}+\mathit{b}+\mathit{c}\right)}^2$
⇒ $3{\mathit{a}}^2+3{\mathit{b}}^2+3{\mathit{c}}^2$ = ${\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2$ + 2ab + 2bc + 2ca
⇒ $2{\mathit{a}}^2+2{\mathit{b}}^2+2{\mathit{c}}^2$ – 2ab – 2bc – 2ca = 0
⇒ ${\mathit{a}}^2+{\mathit{b}}^2$ – 2ab + ${\mathit{b}}^2+{\mathit{c}}^2$ – 2bc + ${\mathit{c}}^2+{\mathit{a}}^2$ – 2ca = 0
⇒ ${\left(\mathit{a}\mathit{–}\mathit{b}\right)}^2+{\left(\mathit{b}\mathit{–}\mathit{c}\right)}^2+{\left(\mathit{c}\mathit{–}\mathit{a}\right)}^2$ = 0
⇒ a – b = 0 ⇒ a = b
[If ${\mathit{x}}^2+{\mathit{y}}^2+{\mathit{z}}^2$ = 0, x = 0, y = 0, z = 0]
b – c = 0 ⇒ b = c
c – a = 0 ⇒ c = a
∴ a = b = c