Three numbers with signs

Test $a=0$. Then $b>0$, and $b\ne0$ gives $c>0$. We would have two positive numbers $b,c$, contradicting the hypothesis. So $a\ne0$.

Test $a>0$. Then $b<0$, and $b\ne0$ gives $c>0$. Again two positives ($a,c$), contradiction. So $a\not>0$; hence $a<0$.

Test $b>0$. Then $b\ne0$ gives $c>0$, another positive besides $b$, contradiction. So $b\le0$; since $a<0$ is the negative number and one number must be zero, $b=0$.

Remaining. $c>0$ is the positive number.

Conclusion. $a<0$, $b=0$, $c>0$.