Question : If $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$, then $a:b:c$ is:
Option 1: $1:1:2$
Option 2: $1:1:1$
Option 3: $1:2:1$
Option 4: $2:1:1$
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Correct Answer: $1:1:1$
Solution :
Given: $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$
$a^{2}+b^{2}+c^{2}-ab-bc-ca=\frac{1}{2}[(a-b)^2 + (b-c)^2+(c-a)^2] =0$
So, $a=b=c$
So, $a:b:c=1:1:1$
Hence, the correct answer is $1:1:1$.
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