If the real numbers $a, b, c$ are such that $a^{2} + 4 b^{2} + 16 c^{2} = 48$ and $a b + 4 b c + 2 c a = 24$ , then what is the value of $a^{2} + b^{2} + c^{2} ?$

12

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16

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21

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31

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Solution

The correct option is D $21$
$a^{2} + 4 b^{2} + 16 c^{2} = 48$

$a b + 4 b c + 2 c a = 24$

$2 a^{2} + 8 b^{2} + 32 c^{2} = 96 . . . . . (i)$

$4 a b + 16 b c + 8 a c = 96 . . . . (i i)$

From equation (i) and (ii), we get

$(- 2 b)^{2} + (2 b - 4 c)^{2} + (4 c - a)^{2} = 0$

$\Rightarrow a = 2 b = 4 c$

$4 b^{2} + 4 b^{2} + 4 b^{2} = 48$

$\Rightarrow 12 b^{2} = 48$

$\Rightarrow b = 2$

$a = 4, c = 1$

$a^{2} + b^{2} + c^{2} = 16 + 4 + 1$

$= 21$

Hence, option $(3)$ is correct.

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Similar questions

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