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${ \begin{cases} x^{ 2 }+y^{ 2 }+xy=1(A) \\ x^{ 2 }+z^{ 2 }+xz=3(B) \\ z^{ 2 }+y^{ 2 }+zy=4(C) \end{cases} }\\ (B)-(A)\quad and\quad (C-A)\quad yield:\\ (z-y)(x+y+z)=2\quad (a)\\ (z-x)(x+y+z)=3(b)\\ \\ Dividing\quad (b)\quad by\quad (a)\quad \\ \frac { (z-x) }{ (z-y) } =\quad 3/2\quad \\ which\quad further\quad simplifies\quad to\quad \boxed { z=3y-2x } \\ Now\quad plugging\quad this\quad value\quad back\quad into\quad (a)\quad we\quad obtain:\\ 4{ y }^{ 2 }-5xy+{ x }^{ 2 }=1(c)\\ Now\quad (c)-(A)\Longrightarrow 3{ y }^{ 2 }=6xy\Longrightarrow \boxed { y=2x } \\ Now\quad z=\quad 6x-2x=\boxed { 4x } \\ Plugging\quad into\quad (b)\\ \boxed { { x }^{ 2 }=\frac { 1 }{ 7 } } \Longrightarrow x=\frac { 1 }{ \sqrt { 7 } } ,y=\frac { 2 }{ \sqrt { 7 } } ,z=\frac { 4 }{ \sqrt { 7 } } \\ Checking\quad if\quad our\quad solution\quad is\quad valid,\\ { \frac { 1+4+2 }{ 7 } }=1(A)\\ \frac { 1+16+4 }{ 7 } =3(B)\\ \frac { 4+16+8 }{ 7 } =4(C)\\ Thus\quad ({ x+y+z) }^{ 2 }=(49{ x }^{ 2 })=\boxed { 7 } \\ \\ \\ \\$