若x/3=y/1=z/2,且xy+yz+zx=99,则2x^2+12y^2+9z^2=

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若x/3=y/1=z/2,且xy+yz+zx=99,则2x^2+12y^2+9z^2=
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若x/3=y/1=z/2,且xy+yz+zx=99,则2x^2+12y^2+9z^2=
若x/3=y/1=z/2,且xy+yz+zx=99,则2x^2+12y^2+9z^2=

若x/3=y/1=z/2,且xy+yz+zx=99,则2x^2+12y^2+9z^2=
令X=3K Y=K Z=2K 代入得：3K^2+2K^2+6K^2=99 所以K=3或-3,故X^2=81 Y^2=9 Z^2=36 所以原式=594

用代数法：
z＝2y，x＝3y代入xy+yz+zx＝99得：3Y^2+2Y^2+6Y^2=99
解得：y^2=9,所以：Z^2=4Y^2=36;X^2=9Y^2=81
所以:2x^2+12y^2+9z^2=162+108+324=594.

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