If $x, y, z$ are real numbers then the range of $x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 x z - 2 x y$ is

ϕ

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R

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[0, \infty)

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(- \infty, 0)

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Solution

The correct option is C $[0, \infty)$
We have,
$x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 x z - 2 x y$
$= x^{2} + (2 y)^{2} + (3 z)^{2} - (x) (2 y) - (2 y) (3 z) - (3 z) (x)$
$= \frac{1}{2} [2 x^{2} + 2 (2 y)^{2} + 2 (3 z)^{2} - 2 (x) (2 y) - 2 (2 y) (3 z) - 2 (3 z) (x)]$
$= \frac{1}{2} [(x - 2 y)^{2} + (2 y - 3 z)^{2} + (x - 3 z)^{2}]$
$(x - 2 y)^{2}, (2 y - 3 z)^{2}, (x - 3 z)^{2}$ are positive as square of real numbers are always positive. So,
$x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 x z - 2 x y$ is always positive
Therefore, range is $[0, \infty)$

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