Given: $X + Y + Z = 37$ and $X/Y = Y/Z \Rightarrow Y^2 = XZ$.

Because 37 is odd, Y must be odd (since XZ has same parity as Y²). Many integer triples satisfy $Y^2 = XZ$ with $X + Y + Z = 37$. For example:

• Let $Y = 12$ (not allowed as Y must be odd) — adjust to an odd Y: Take $Y = 11 \Rightarrow Y^2 = 121$. Choose $(X, Z) = (11, 11)$: sums to 33 (too small). Choose $(X, Z) = (9, 121/9)$ not integer.

A concrete working pair with the same ordering issue can be built by scaling a geometric progression: Let $X = a$, $Y = ar$, $Z = ar^2$ with $r = Y/X = Z/Y$ and integers summing to 37.

Example 1: $(X, Y, Z) = (9, 12, 16)$ gives $Y^2 = 144 = 9 \times 16$ and sum 37. Ordering: $X < Y < Z$.

Example 2: Swap X and Z: $(X, Y, Z) = (16, 12, 9)$ also satisfies $Y^2 = 144 = 16 \times 9$ and sum 37. Ordering: $Z < Y < X$.

Since both orderings are possible with the same constraints, the relative order of Value-I, Value-II, and Value-III cannot be uniquely determined.

Therefore, option D is correct.