Condition $\gcd(x, y)=\gcd(y, z)=\gcd(z, x)=1$ gives us: $x|y+z$, $y|x+z$, $z|x+y$ and hence $xyz|x+y+z$ from which we easily get $(1,1,1)$ and $(1,1,2)$ and permutations as the only solution (and $t$ can be easily determined)

(x,y,z)=(1,2,3)?

Very well remarked, yunxiu! Can you provide a correct, complete solution?

From $ k=\frac{x+y+z}{xyz}\in \mathbb{Z}_+$ , WLOG $ x\le y \le z$ then $ x+y+z\le 3z$ then $ k \le \frac{3z}{xyz}$ then $ xy \le \frac{3}{k}$ then $ xy \le 3$ then possible solutions are: $ (x,y)=(1,1)$ with $ k=\frac{2+z}{z}$ with $ (x,y,z)=(1,1,1)$ or $ (1,1,2)$ $ (x,y)=(1,2)$ with $ k=\frac{3+z}{z}$ with $ (x,y,z)=(1,2,3)$ $ (x,y)=(1,3)$ with $ k=\frac{4+z}{z}$ with candidate $ (x,y,z)=(1,3,4)$ however does not fullfilled for $ y$

so nice !!