1. Statement of the problem. We seek all positive integers $x,y,z$ such that $x+y+z$ is prime and $xy+yz+zx$ divides $x^2+y^2+z^2$. 2. Notation and identity. Let $p=x+y+z$ which is prime. Using the identity $$x^2+y^2+z^2 = p^2 - 2(xy+yz+zx).$$ Because $xy+yz+zx$ divides $x^2+y^2+z^2$ it follows that $xy+yz+zx$ divides $p^2$. Hence the positive integer $xy+yz+zx$ must be a divisor of $p^2$ so $xy+yz+zx\in \{1,p,p^2\}$. 3. Exclude impossible divisors. Note $xy+yz+zx\ge3$ for positive integers $x,y,z$ so it cannot equal 1. If $xy+yz+zx = p^2$ then by the identity above $$x^2+y^2+z^2 = p^2 - 2p^2 = -p^2<0$$ which is impossible. Thus $xy+yz+zx\neq p^2$. 4. Analyze $xy+yz+zx=p$. Write $x=1+a$, $y=1+b$, $z=1+c$ with integers $a,b,c\ge0$. Then $p=3+a+b+c$. Compute $$xy+yz+zx = (1+a)(1+b)+(1+b)(1+c)+(1+c)(1+a) = ab+bc+ca+2(a+b+c)+3.$$ Subtracting $p$ gives $$xy+yz+zx - p = ab+bc+ca+(a+b+c)\ge0.$$ Therefore $xy+yz+zx\ge p$ with equality exactly when $a=b=c=0$, i.e. when $x=y=z=1$. When $xy+yz+zx=p$ we have $$\frac{p^2}{p} = \frac{\cancel{p}p}{\cancel{p}} = p$$ which is consistent. 5. Conclusion. The only possibility that satisfies all conditions is $x=y=z=1$ and then $x+y+z=3$ is prime and $xy+yz+zx=3$ divides $x^2+y^2+z^2=3$. Thus all positive integer triples are $(1,1,1)$.