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

[0, \infty)

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(- \infty, 0)

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(1, \infty)

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(2, \infty)

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Solution

The correct option is B $[0, \infty)$
The given expression $x^{2} + 4 y^{2} + 9 z^{2} - 6 y z + 3 z x - 2 x y$ can be written as $\frac{(x - 2 y)^{2} + (2 y - 3 z)^{2} + (3 z + x)^{2}}{2}$ , which is the sum of $3$ square terms. Hence, always positive.
So, the range is $[0, \infty)$ .

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