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

Answer: Option C

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