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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there is no limit to least value.

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0

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none of these

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Solution

The correct option is A $1$
Given $x^{2} + 4 y^{2} + 3 z^{2} - 2 x - 12 y - 6 z + 14$

$= x^{2} - 2 x + 1 + {4 y}^{2} + 9 - 12 y + 3 z^{2} + 3 - 6 z + 1$

$= {(x - 1)}^{2} + 4 {(y - \frac{3}{2})}^{2} + 3 {(z - 1)}^{2} + 1 \geq 1$

the sum of squares is always $\geq 0$

$\Rightarrow m i n ({(x - 1)}^{2} + 4 {(y - \frac{3}{2})}^{2} + 3 {(z - 1)}^{2}) = 0$

$∴ min (x^{2} + 4 y^{2} + 3 z^{2} - 2 x - 12 y - 6 z + 14) = 0 + 1 = 1$

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