Solving the following equations : $x^{2}+xy+y^{2}=37,\ y^{2}+yz+z^{2}=19,\ z^{2}+zx+z^{2}=28.$ we get $x=4,-4,\dfrac { 10\sqrt { 3 } }{ 3 } ,-\dfrac { 10\sqrt { 3 } }{ 3 },$ $y=3,-3\dfrac { 1 }{ 3 } \sqrt { 3 } ,-\dfrac { 1 }{ 3 } \sqrt { 3 },$ $z=2,-2,-\dfrac { 8 }{ 3 } \sqrt { 3 } ,\dfrac { 8 }{ 3 } \sqrt { 3 }$.

To verify the solutions, we substitute the given positive values x = 4, y = 3, and z = 2 into the provided equations. For the first equation x^2 + xy + y^2, calculating 4^2 + 4*3 + 3^2 gives 16 + 12 + 9, which equals 37. For the second equation y^2 + yz + z^2, calculating 3^2 + 3*2 + 2^2 gives 9 + 6 + 4, which equals 19. For the third equation z^2 + zx + x^2, calculating 2^2 + 2*4 + 4^2 gives 4 + 8 + 16, which equals 28. Since the values satisfy all three equations, the statement is true.

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