If a² + b² + 4c² = 2(a + b - 2c) - 3 and a, b, c are real, then the value of (a² + b² + c²) is?

Solution (By Examveda Team)

$$\eqalign{ & {a^2} + {b^2} + 4{c^2} = 2\left( {a + b - 2c} \right) - 3 \cr & \Rightarrow {a^2} + {b^2} + 4{c^2} - 2a - 2b + 4c + 3 = 0 \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} - 2b + 1 + 4{c^2} + 4c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {2c + 1} \right)^2} = 0 \cr & \cr & \therefore a - 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 1 \cr & \,\,\,\,\,\,\,b - 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b = 1 \cr & \,\,\,\,2c + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c = \frac{{ - 1}}{2} \cr & \cr & \therefore {a^2} + {b^2} + {c^2} \cr & \Rightarrow 1 + 1 + \frac{1}{4} \cr & \Rightarrow 2 + \frac{1}{4} \cr & \Rightarrow \frac{9}{4} \cr & \Rightarrow 2\frac{1}{4} \cr} $$