A) Non-negative
B) Non-positive
C) Zero
D) None of these
Correct Answer: A
Solution :
\[x,y,z\in R\]and distinct. Now, \[u={{x}^{2}}+4{{y}^{2}}+9{{z}^{2}}-6yz-3zx-2xy\] \[=\frac{1}{2}(2{{x}^{2}}+8{{y}^{2}}+18{{z}^{2}}-12yz-6zx-4xy)\] \[=\frac{1}{2}\left\{ {{x}^{2}}-4xy+4{{y}^{2}})+({{x}^{2}}-6zx+9{{z}^{2}})+(4{{y}^{2}}-12yz+9{{z}^{2}}) \right\}\] \[=\frac{1}{2}\left\{ {{(x-2y)}^{2}}+{{(x-3z)}^{2}}+{{(2y-3z)}^{2}} \right\}\] Since it is sum of squares. So \[u\] is always non- negative.
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