If x, y, z are real and distinct,then u = x^2 | vedclass

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(A) Given $u = x^2 + 4y^2 + 9z^2 - 6yz - 3zx - 2xy$.Multiply and divide by $2$:$u = \frac{1}{2}(2x^2 + 8y^2 + 18z^2 - 12yz - 6zx - 4xy)$Rearrange the

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Non-negative

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(A) Given $u = x^2 + 4y^2 + 9z^2 - 6yz - 3zx - 2xy$.Multiply and divide by $2$:$u = \frac{1}{2}(2x^2 + 8y^2 + 18z^2 - 12yz - 6zx - 4xy)$Rearrange the terms to form perfect squares:$u = \frac{1}{2}[(x^2 - 4xy + 4y^2) + (x^2 - 6zx + 9z^2) + (4y^2 - 12yz + 9z^2)]$Simplify the expressions inside the brackets:$u = \frac{1}{2}[(x - 2y)^2 + (x - 3z)^2 + (2y - 3z)^2]$Since $x, y, z$ are real,the sum of squares is always non-negative. Thus,$u \geq 0$.

If $x, y, z$ are real and distinct,then $u = x^2 + 4y^2 + 9z^2 - 6yz - 3zx - 2xy$ is always

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