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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