The least value of the expression $x^{2} + 4 y^{2} + 3 z^{2} - 2 x - 12 y - 6 z + 14$ is

1

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0

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No least value

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None of thses

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Solution

The correct option is C $0$
Now
$x^{2} + 4 y^{2} + 3 z^{2} - 2 x - 12 y - 6 z + 19$

$= (x^{2} - 2 x + 1) + (4 y^{2} - 12 y + 9) + 3 (z^{2} - 2 z + 1)$

$= {(x - 1)}^{2} + {(2 y - 3)}^{2} + 3 {(z - 1)}^{2}$

The given expression is sum of three square terms and

${(x - 1)}^{2} \geq 0$ , ${(2 y - 3)}^{2} \geq 0$ and ${(z - 1)}^{2} \geq 0$

The least value of the expression is $0$

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