Note that $ab\equiv bc\equiv ca\equiv -1 \mod p \implies a\equiv b\equiv c\equiv \frac{a+b+c}{3}\mod p$ and $a^2\equiv b^2\equiv c^2 \equiv -1 \mod p$ assume the contrary that $\frac{a+b+c}{3}\le p+1$ clearly $\frac{a+b+c}{3}<p$ So $r=\frac{a+b+c}{3}$ is the remainder of $a,b,c$ when divided by $p$ So : $a=q_1p+r,b=q_2p+r,c=q_3p+r \implies q_1+q_2+q_3=0\implies q_1=q_2=q_3=0$ a contradiction So the inequality is true now assume $\frac{a+b+c}{3}=p+2\implies a\equiv b\equiv c\equiv \frac{a+b+c}{3} \equiv 2 \mod p$ So $ab+1\equiv 5\mod p\implies p=5$ An example is $(a,b,c)=(2,5,12)$