If $a^{2} + 10 b^{2} + 5 c^{2} + 6 a b + 2 b c - 16 c + 16 = 0$ then the possible value of $a - b + c$ =_________

5

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- 3

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- 2

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10

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Solution

The correct option is D $10$
Given that,
$a^{2} + 10 b^{2} + 5 c^{2} + 6 a b + 2 b c - 16 c + 16 = 0$

$⟹ (a^{2} + 6 a b + 9 b^{2}) + (b^{2} + 2 b c + c^{2}) + (4 c^{2} - 16 c + 16) = 0$

$⟹ (a + 3 b)^{2} + (b + c)^{2} + (2 c - 4)^{2} = 0$

$[∵ x^{2} + y^{2} + z^{2} = 0 ⟹ x = 0, y = 0, z = 0]$

$⟹ a + 3 b = 0, b + c = 0, 2 c - 4 = 0$

$⟹ a = - 3 b, b = - c, 2 c = 4$

$⟹ a = 6, b = - 2, c = 2$

$∴ a - b + c = 6 - (- 2) + 2 = 6 + 2 + 2 = 10$

Option D is correct.

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