lf $x, y, z$ are real and distinct, then $x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y$ is always

positive

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non negative

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negative

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zero

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Solution

The correct option is D non negative
$x^{2} + 4 y^{2} + 9 z^{2} - 6 y z - 3 z x - 2 x y$
$= \frac{1}{2} ((x - 2 y)^{2} + (2 y - 3 z)^{2} + (3 z - x)^{2})$
This is the sum of 3 squares. So this term is greater than or equal to 0.
Equality occurs when, $x = 2 y$ , $2 y = 3 z$ , $3 z = x$

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