Question : If $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$, then
Option 1: $a=b=c$
Option 2: $a\neq b=c$
Option 3: $a=b\neq c$
Option 4: $a\neq b\neq c$
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Correct Answer: $a=b=c$
Solution :
Given: $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$
Multiplying the whole equation by 2,
$2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca=0$
⇒ $(a-b)^2+(b-c)^2+(c-a)^2=0$
⇒ $(a-b)^2=(b-c)^2=(c-a)^2=0$ [If the sum of squares of three numbers is zero, then each number will also be zero]
⇒ $a=b=c$
Hence, the correct answer is $a=b=c$.
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