If a² + b² + c² + 96 = 8(a + b - 2c), then $$\sqrt {ab - bc - ca} $$   is equal to

Solution (By Examveda Team)

$$\eqalign{ & {a^2} + {b^2} + {c^2} + 96 = 8\left( {a + b - 2c} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 96 = 8a + 8b - 16c \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 96 - 8a - 8b + 16c = 0 \cr & \Rightarrow {a^2} - 8a + 16 + {b^2} - 8b + 16 + {c^2} + 16c + 64 = 0 \cr & \Rightarrow {a^2} - 2 \times a \times 4 + {4^2} + {b^2} - 2 \times b \times 4 + {4^2} + {c^2} + 2 \times c \times 8 + {8^2} = 0 \cr & \Rightarrow {\left( {a - 4} \right)^2} + {\left( {b - 4} \right)^2} + {\left( {c + 8} \right)^2} = 0 \cr & {\text{Now we can say,}} \cr & \Rightarrow {\left( {a - 4} \right)^2} = 0 \cr & \Rightarrow a - 4 = 0 \cr & \Rightarrow a = 4 \cr & {\text{Similarly,}} \cr & \Rightarrow {\left( {b - 4} \right)^2} = 0 \cr & \Rightarrow b - 4 = 0 \cr & \Rightarrow b = 4 \cr & {\text{Similarly,}} \cr & \Rightarrow {\left( {c + 8} \right)^2} = 0 \cr & \Rightarrow c + 8 = 0 \cr & \Rightarrow c = - 8 \cr & {\text{Now, put }}a,{\text{ }}b{\text{ and }}c{\text{ value in }}\sqrt {ab - bc - ca} \cr & \Rightarrow \sqrt {4 \times 4 - 4 \times \left( { - 8} \right) + \left( { - 8} \right) \times 4} \cr & \Rightarrow \sqrt {16 + 32 - 32} \cr & \Rightarrow \sqrt {16} \cr & \Rightarrow 4 \cr} $$