Given : a2+b2+c2=2(a−b−c)−3 ⇒ a2+b2+c2=2a−2b−2c−3 ⇒ (a2−2a)+(b2+2b)+(c2+2c)=−(1+1+1) ⇒ (a2−2a+1)+(b2+2b+1)+(c2+2c+1)=0 ⇒ (a−1)2+(b+1)2+(c+1)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