Saudi Arabian Mathematical Competitions

From the given conditions, we have
$$xz = 2z - y, \quad xy = 2x - z, \quad yz = 2y - x.$$
Taking the sum of these equations, side by side, we have
$$xy + yz + zx = x + y + z = T.$$
From $xz = 2z - y$, we also can get $2z - x - y = xz - x \rightarrow 3z - T = x(z - 1)$. Make the similar equations and multiply them, we get
$$xyz(x - 1)(y - 1)(z - 1) = (3x - T)(3y - T)(3z - T).$$
Note that $(x - 1)(y - 1)(z - 1) = xyz - (xy + yz + zx) + (x + y + z) - 1 = xyz - 1$ and
$$\begin{aligned} (3x - T)(3y - T)(3z - T) &= 27xyz - 9T(xy + yz + zx) + T^2(3x + 3y + 3z) - T^3 \\ &= 27xyz - 9T^2 + 2T^3 \end{aligned}$$
Thus
$$\begin{equation*} xyz(xyz - 1) = 27xyz - 9T^2 + 2T^3 \text{ or } (xyz)^2 - 28xyz = 2T^3 - 9T^2. \tag{1} \end{equation*}$$
By the similar transformation, $xz = 2z + y \rightarrow xz + 2x = 2z + 2x - y \rightarrow x(z + 2) = 2T - 3y$. Make the similar equations and multiply them, side by side, we get
$$(xyz)^2 + 6(xyz)T + 35xyz = 18T^2 - 4T^3.$$
On the other hand, $xz = 2z - y \rightarrow xyz = 2yz - y^2$. Make the similar equations and multiply them, side by side, we get
$3xyz = 2(xy + yz + zx) - (x^2 + y^2 + z^2) = 4(xy + yz + zx) - (x + y + z)^2 = 4T - T^2$.
So we have $xyz = \frac{4T - T^2}{3}$, then by substituting to equation (1), we get
$$\begin{equation*} \left(\frac{4T - T^2}{3}\right)^2 - 28\left(\frac{4T - T^2}{3}\right) = 2T^3 - 9T^2 \tag{2} \end{equation*}$$
Solving this equation, we have $T \in \{0, 3, 7, 16\}$.

1. If $T = 0$, then $xyz = 0$, a contradiction.
2. If $T = 16$, then $xyz = -64$ and this pair $(T, xyz) = (16, -64)$ does not satisfy equation (2).
3. If $T = 3$, then $xyz = 1$ and by Vieta's theorem, $x, y, z$ are real roots of $t^3 - 3t^2 + 3t - 1 = 0$, which implies that $x = y = z = 1$.
4. If $T = 7$, we get $xyz = -7$ and $x, y, z$ are roots of $t^3 - 7t^2 + 7t + 7 = 0$, this also have three distinct real roots.

Hence, all possible values of $T$ are $3$ or $7$.