If a² + b² + c² = 2(a - b - c) -3, then the value of a + b + c is?

Answer:- 1

Explanation:-

Solution:

\begin{aligned} a^{2} + b^{2} + c^{2} = 2 (a - b - c) - 3 \\ \Rightarrow a^{2} + b^{2} + c^{2} = 2 a - 2 b - 2 c - 3 \\ \Rightarrow a^{2} + b^{2} + c^{2} - 2 a + 2 b + 2 c + 3 = 0 \\ \Rightarrow a^{2} - 2 a + 1 + b^{2} + 2 b + 1 + c^{2} + 2 c + 1 = 0 \\ \Rightarrow {(a - 1)}^{2} + {(b + 1)}^{2} + {(c + 1)}^{2} = 0 \\ {(a - 1)}^{2} = 0 {(b + 1)}^{2} = 0 {(c + 1)}^{2} = 0 \\ \Rightarrow a - 1 = 0 \Rightarrow b + 1 = 0 \Rightarrow c + 1 = 0 \\ \Rightarrow a = 1 \Rightarrow b = - 1 \Rightarrow c = - 1 \\ ∴ a + b + c \\ = 1 - 1 - 1 \\ = - 1 \end{aligned}

$\eqalign{ & {a^2} + {b^2} + {c^2} = 2\left( {a - b - c} \right) - 3 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 2a - 2b - 2c - 3 \cr & \Rightarrow {a^2} + {b^2} + {c^2} - 2a + 2b + 2c + 3 = 0 \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} + 2b + 1 + {c^2} + 2c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b + 1} \right)^2} + {\left( {c + 1} \right)^2} = 0 \cr & {\left( {a - 1} \right)^2} = 0{\text{ }}{\left( {b + 1} \right)^2} = 0{\text{ }}{\left( {c + 1} \right)^2} = 0 \cr & \Rightarrow a - 1 = 0{\text{ }} \Rightarrow b + 1 = 0{\text{ }} \Rightarrow c + 1 = 0 \cr & \Rightarrow a = 1{\text{ }} \Rightarrow b = - 1{\text{ }} \Rightarrow c = - 1 \cr & \therefore a + b + c \cr & = 1 - 1 - 1 \cr & = - 1 \cr}$