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1 solution
\frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } =\frac { xy+yz+zx }{ xyz } =0\quad \therefore \quad xy+yz+zx=0\\ \\ \Rightarrow \quad ({ x+y+z }^{ 2 })={ x }^{ 2 }{ +y }^{ 2 }+{ z }^{ 2 }+2(xy+yz+zx)\\ \Rightarrow \quad x+y+z=\pm \sqrt { 2+0 } \\ \Rightarrow x+y+z=\pm \sqrt [ 2 ]{ 2 } \\ \Rightarrow ({ x+y+z) }^{ 12 }={ \sqrt [ 2 ]{ \pm 2 } }^{ 12 } =64