We multiply by $4$ and rewrite the equation as $(2x-1)^{2}+(2y-1)^{2}+(2z-1)^{2}=7$. As the numbers $2x-1, 2y-1, 2z-1$ are rational (because $x,y,z$ are) we can multiply by all the denominators (squared) and get to the equation $a^{2}+b^{2}+c^{2}=7d^{2}$ in whole numbers. Now if $d=0$, then one of the denominators is $0$, which is impossible. Now if $2\vert d$, then $d \equiv 0,2$ (mod $4$) we get $2\vert a, 2\vert b, 2\vert c$ and we can set $a=2a_{1}, b=2b_{1}, c=2c_{1}, d=2d_{1}$ and look at the same equation for the variables $a_{1}, b_{1}, c_{1}, d_{1}$. As $d$ is not a zero, $v_{2}(d)$ is finite so we can assume that $d$ is odd. Now $RHS \equiv 7$ (mod $8$), but $LHS \equiv 0,1,2,3$ (mod $8$), which is a contradiction and we are done as $RHS\neq LHS$.