Let $\gcd (x,y,z)=d$ then $x=dx_1,y=dy_1,z=dz_1$ with $\gcd (x_1,y_1,z_1)=1$. This implies $\gcd (x_1+y_1,y_1+z_1,z_1+x_1)=k>1$. We have $k \mid (x_1+y_1)+(y_1+z_1)$ and $k \mid x_1+z_1$ so $k \mid 2y_1$. Similarly $k \mid 2z_1,k \mid 2x_1$. If there exists $2 \le l \mid k$ such that $l \mid x_1$ then from $k \mid x_1+y_1,z_1+x_1$ we follows $l \mid y_1,l \mid z_1$, a contradiction since $\gcd (x_1,y_1,z_1)=1$. Hence, $\gcd (k,x_1)=1$ implies $k \mid 2$. Since $k >1$ so $k=2$. We follows that $2 \nmid x_1y_1z_1$. The answer is $(x,y,z)=(dx_1,dy_1,dz_1)$ with $\gcd (x_1,y_1,z_1)=1$ and $2 \nmid x_1y_1z_1$.