If $a^{2} + 2 b = 7, b^{2} + 4 c = - 7$ and $c^{2} + 6 a = - 14,$ then the value of $(a^{2} + b^{2} + c^{2})$ is

14

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25

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36

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47

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Solution

The correct option is A $14$
Given $a^{2} + 2 a = 7, b^{2} + 4 c = - 7$ and $c^{2} + 6 a = - 14$
Then, $a^{2} = 7 - 2 b$ , $b^{2} = - 7 - 4 c$ , $c^{2} = - 14 - 6 a$
Add three equation then
$a^{2} = 7 - 2 b$ , $b^{2} = - 7 - 4 c$ , $c^{2} = - 14 - 6 a$
$\Rightarrow a^{2} + b^{2} + c^{2} + 6 a + 2 b + 4 c + 14 = 0 \Rightarrow a^{2} + 6 a + 9 + b^{2} + 2 b + 1 + c^{2} + 4 c + 4 = 0$
$\Rightarrow (a + 3)^{2} + (b + 1)^{2} + (c + 2)^{2} = 0$
Hence $(a + 3)^{2} = 0 \Rightarrow a + 3 = 0 \Rightarrow a = - 3$
Or $(b + 1)^{2} = 0$

$\Rightarrow b + 1 = 0$ and $b = - 1$

Also $(c + 2)^{2} = 0 \Rightarrow c + 2 = 0 \Rightarrow c = - 2$
Then, $a^{2} + b^{2} + c^{2}$
$= (- 3)^{2} + (- 1)^{2} + (- 2)^{2} = 9 + 1 + 4 = 14$

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