Question : If $a^2+b^2+c^2+216=12(a+b-2 c)$, then $\sqrt{a b-b c+c a}$ is:
Option 1: 8
Option 2: 3
Option 3: 6
Option 4: 4
Correct Answer: 6
Solution :
Given: $a^2+b^2+c^2+216=12(a+b-2 c)$
Rearranging and simplifying
⇒ $a^2 - 12a + b^2 - 12b + c^2 + 24c + 216 = 0$
⇒ $a^2 - 12a + 36 + b^2 - 12b + 36 + c^2 + 24c + 144 = 0$
⇒ $(a - 6)^2 + (b - 6)^2 + (c + 12)^2 = 0$
$\therefore a = 6, b = 6,$ and $c = -12$
Now,
$\sqrt{a b-b c+c a}$
$=\sqrt{6 \times 6 -(6 \times (-12))+ (-12) \times 6}$
$=\sqrt{36 + 72 - 72}$
$= \sqrt{36} = 6$
Hence, the correct answer is 6.
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