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Question

If x2+y2+z264xyyzzx=2 and x+y=3z, then the value of z is

A
2
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B
3
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C
4
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D
2
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Solution

The correct option is C 4
x2+y2+z264xyyzzx=2
x2+y2+z2=2xy+2yz+2zx+64
Now, (x+y+z)2=x2+y2+z2+2xy+2yz+2zx
(3z+z)2=2xy+2yz+2zx+64+2xy+2yz+2zx
16z2=4yz+4xz+64
16z2=4z(x+y)+64
16z2=4z(3z)+64
16z212z2=64
4z2=64
z2=16
z=4

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