由算术平方根的意义知：x+y-2002≥0且2002-x-y≥0
解之,得：x+y=2002
于是,根号3x+y-z-2 +根号2x+y-z = 0
可得：3x+y-z-2=0且2x+y-z = 0
故,x=2, y=2000, z=2004.
x y z适合关系式√3x+y-z-2+√2x+y-2002+√2002-x-y,求x y z的值
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