Is it geometry?

Consider x,y,z to be the sides of a right triangle with z as hypotenuse ... 2nd condition implies that angle between sides z and y =60 degrees (cosine rule).

Now Apply cosine cosine rule to $y^2$ i.e

$y^2 = x^2 + z^2 - \sqrt{3} xz$ . So k = -1.732

We can replace $z^{2}$ in second equation from first equation. $y^{2}-yz+ x^{2} + y^{2}=x^{2}$ $y(2y-z)=0$ Since $x,y,z$ are $+ve$ numbers, therefore $y\ne 0$ . $z=2y$ $y = \dfrac{z}{2}$ Again we can replace $z^{2}$ in third equation from first equation. $x^{2}+y^{2} + kzx+x^{2}=y^{2}$ $x(2x+kz)=0$ Again with the same logic as before, we can claim that (x\ne 0). $x = \dfrac{-kz}{2}$ We can replace these new found values of $x$ and $y$ in first equaion. $\dfrac{z^{2}}{2} + \dfrac{k^{2}z^{2}}{4} = z^{2}$ $1+ k^{2} = 4$ $k=\pm \sqrt{3}$ But $k$ has to $-ve$ as $x$ is $+ve$ . Hence $k=-\sqrt{3}$ .

$\begin{cases} { x }^{ 2 }+{ y }^{ 2 }={ z }^{ 2 }....(1) \\ { y }^{ 2 }-yz+{ z }^{ 2 }={ x }^{ 2 }....(2) \\ { z }^{ 2 }+k·zx+{ x }^{ 2 }={ y }^{ 2 }....(3) \end{cases}\\ Substituting~z~from~(1)~in~(2)~we~get~~~2y^2=yz,~~~\implies~z=2y, ....(4)~~y\neq 0.\\ From~(4)~and~(1) ~x^2=3y^2....(5)\\ Substituting~y~and~z~from~(5)~and~(4)~in~(3)~we~get,\\ 4y^2 \pm 2\sqrt3*k*y^2+3y^2=y^2....(6) \\ \implies~3y^2=\pm \sqrt3*k*y^2.\\ \therefore~k=\pm \sqrt3.\\ Puting~ these~values~in~ (6), ~~we~see~that~k=\Large \color{#D61F06}{ - \sqrt3 = - 1.732}.$ )