Answers

2014-04-24T16:53:22+05:30
There is an identity where (x+y+z)=x^2+y^2+z^2 +2xy+2yz+2zy
substituting the values
(10)^2=40+2(x^2+y^2+z^2 +2xy+2yz+2zy)
60=2(x^2+y^2+z^2 +2xy+2yz+2zy)
(x^2+y^2+z^2 +2xy+2yz+2zy)=30
option B is correct

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