Question : If $9 a+9 b+9 c=81$ and $4 a b+4 b c+4 c a=160$, then what is the value of $6 a^2+6 b^2+6 c^2$?
Option 1: 1
Option 2: 3
Option 3: 4
Option 4: 6
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Correct Answer: 6
Solution : $9a + 9b + 9c = 81$ ⇒ $a + b + c = 9$.............(i) $4ab + 4bc + 4ca = 160$ ⇒ $ab + bc + ca = 40$..................(ii) We know that, $(a + b + c)^2 = a^2+ b^2 + c^2 + 2ab + 2bc + 2ca$ Putting the values from the above equations, we get, ⇒ $9^2 = a^2 + b^2 + c^2 + 2 × 40$ ⇒ $a^2 + b^2 + c^2 = 81 - 80 = 1$ Multiplying both sides by 6, we get, $\therefore6a^2 + 6b^2 + 6c^2 = 6$ Hence, the correct answer is 6.
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