Question : If $(a+b+c)=17$ and $\left(a^2+b^2+c^2\right)=101$, then find the value of $(a-b)^2+(b-c)^2+(c-a)^2$.
Option 1: 12
Option 2: 14
Option 3: 10
Option 4: 16
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Correct Answer: 14
Solution : Given: $(a+b+c)=17$ and $\left(a^2+b^2+c^2\right)=101$ We know that $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$ Squaring both sides, we get ⇒ $a^2+b^2+c^2+2ab+2bc+2ac=289$ ⇒ $101+2ab+2bc+2ac=289$ ⇒ $2ab+2bc+2ac=188$ Now, $(a-b)^2+(b-c)^2+(c-a)^2$ = $2a^2+2b^2+2c^2-2ab-2bc-2ac$ = $2\times101-188$ = $202-188=14$ Hence, the correct answer is 14.
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