Question : If $a^2+b^2+c^2=14$ and $a+b+c=6$, then the value of $(ab+bc+ca)$ is:
Option 1: 11
Option 2: 12
Option 3: 13
Option 4: 14
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Correct Answer: 11
Solution : Given: $a^2+b^2+c^2=14$ and $a+b+c=6$. On squaring the given equation $a+b+c=6$, we get, $(a+b+c)^2=6^2$ ⇒ $a^2+b^2+c^2+2(ab+bc+ca)=36$ Substitute the value of $a^2+b^2+c^2=14$ in the above equation and we get, ⇒ $14+2(ab+bc+ca)=36$ ⇒ $2(ab+bc+ca)=22$ ⇒ $(ab+bc+ca)=11$ Hence, the correct answer is 11.
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