Positive integers in AP product condition

Let $x = y - d$, $z = y + d$, where $d > 2$ is the common difference. Then $z - x = 2d$.

Sum $x + y + z = 3y$. $xyz = (y - d) y (y + d) = y(y^2 - d^2)$.

Given equation: $y(y^2 - d^2) = 5(3y) = 15y$ Since $y > 0$, divide by $y$: $y^2 - d^2 = 15 \implies (y - d)(y + d) = 15$

Since $x = y - d$ and $z = y + d$ are positive integers and $d > 2$: Factors of 15 are $(1, 15)$ and $(3, 5)$.

Case 1: $y - d = 1$ and $y + d = 15$ Adding gives $2y = 16 \implies y = 8$, so $d = 7$. Check $d > 2$: $7 > 2$ holds. Here $z - x = 2d = 14$.

Case 2: $y - d = 3$ and $y + d = 5$ Adding gives $2y = 8 \implies y = 4$, so $d = 1$. Check $d > 2$: $1 > 2$ fails.

Thus $z - x = 14$.