If $a^2+b^2+c^2+216=12(a+b-2 c)$, then $\sqrt{a b-b c-c a}$ is:
Answer & explanation
Correct answer: option 3
(a + b)2 = a2 + b2 + 2ab
(a – b)2 = a2 + b2 – 2ab
a2 + b2 + c2 + 216 = 12(a + b – 2c)
a2 + b2 + c2 + 216 = 12(a + b – 2c)
= a2 – 12a + b2 – 12b + c2 + 24c + 216 = 0
= a2 – 12a + 36 + b2 – 12b + 36 + c2 + 24c + 144 = 0
= (a – 6)2 + (b – 6)2 + (c + 12)2 = 0
a = 6
b = 6
c = –12
$\sqrt{a b-b c-c a}$ = $\sqrt{6 × 6 - 6 × -12 - (-12) × 6}$
$\sqrt{a b-b c-c a}$ = $\sqrt{180}$ = $6 \sqrt{5}$