Lemma Let $ ( x,y,c) \in \mathbb{R}^{3}_{\geq 0}$ such that $ x+y\leq 2c$. Then we have $ xy\leq c^{2}$. This is clear from $ 4xy=( x+y)^{2} -( x-y)^{2} \leq ( x+y)^{2}$. $ 1/a+1/b+1/c=1/p\ \cdots ( \heartsuit )$ $ a+b+c< 2p\sqrt{p} \ \ \cdots ( \clubsuit )$ From $ ( \heartsuit )$, we have $ a,b,c\geq p+1\ \cdots ( \eighthnote )$ First, we wiil show that $ \gcd( a,b,c)$ is divisible by $ p$. Assume that $ ( a,b,c)$ satisfies $ ( \heartsuit )$,$ ( \clubsuit )$, and $ p\nmid \gcd( a,b,c)$. From $ ( \heartsuit )$, we have $ p( ab+bc+ca) =abc\ \cdots ( \diamondsuit )$. WLOG, we can take $ s\in \mathbb{N}$ such that $ c=ps$. From $ ( \diamondsuit )$ and $ c=ps$, we have $ ps( a+b) =( s-1) ab\ \cdots ( \spadesuit )$. We can show that $ p\nmid s-1$ as follows : Assume that $ p\mid s-1$, then $ s\geq p+1$. So we have $ c=ps\geq p( p+1)$. From $ c\geq p( p+1)$ and $ ( \eighthnote )$, $ a+b+c\geq p^{2} +3p+2$, which contradicts $ ( \clubsuit )$. Since $ p\nmid s-1$ and $ ( \spadesuit )$, wlog we can take $ t\in \mathbb{N}$ such that $ b=pt$. From $ b=pt$ and $ ( \spadesuit )$, we have $ stp=( st-s-t) a\ \cdots ( \twonotes )$. From $ p\nmid \gcd( a,b,c)$ and $ ( \twonotes )$, we must have $ p\mid st-s-t$. In particular, we have $ p\leq st-s-t< st\ \cdots ( 1)$. From $ b=pt$, $ c=ps$, and $ ( \clubsuit )$, we have $ s+t< 2\sqrt{p} \ \cdots ( 2)$. From $ ( 2)$ and lemma, we must have $ st\leq p$, which contradicts $ ( 1)$. So we must have $ p\mid \gcd( a,b,c)$. So we can take $ ( x,y,z) \in \mathbb{N}^{3}$ such that $ ( a,b,c) =( px,py,pz)$. Thus we have $ 1/x+1/y+1/z=1\ \cdots ( 3)$. From $ ( \clubsuit )$, we have $ x+y+z< 2\sqrt{p} \ \cdots ( 4)$. WLOG, $ x\geq y\geq z$. If $ z\geq 3$, then $ 1/x+1/y+1/z\leq 1/3+1/3+1/3$. So we have $ x=y=z=3$, and $ ( x,y,z) =( 3,3,3)$ is a solution for $ ( 3)$. If $ z=2$, then $ 1/x+1/y=1/2$, which implise $ ( x-2)( y-2) =4$. It follows that $ ( x,y) =( 6,3) ,( 4,4)$. Thus $ ( x,y,z) =( 3,3,3) ,( 6,3,2) ,( 4,4,2) \ \cdots ( 5)$. From $ ( 4)$ and $ ( 5)$, we have the following conclusion : If $ p< 23$, then there is no solution. If $ p=23$, then $ ( a,b,c) =( 3p,3p,3p)$. If $ p=29$, then $ ( a,b,c) =( 3p,3p,3p\} ,( 4p,4p,2p) ,( 4p,2p,4p) ,( 2p,4p,4p)$. If $ p\geq 31$, then $ ( a,b,c) =( 3p,3p,3p\} ,( 4p,4p,2p) ,( 4p,2p,4p) ,( 2p,4p,4p)$ $ ( 6p,3p,2p) ,( 6p,2p,3p) ,( 2p,3p,6p) ,( 2p,6p,3p) ,( 3p,2p,6p) .( 3p,6p,2p)$.