In the $xyz$ space, prove that the sphere $S$ centered the origin $O$ with radius $\sqrt{6}$ has common points with the plane $\alpha$ passing through 3 points $(4,\ 0,\ 0),\ (0,\ 4,\ 0),\ (0,\ 0,\ 4)$. Then when the point $(x,\ y,\ z)$ moves on the whole set of the common points, find the range of $xyz$.

[i]2011 Kyoto University enrance exam/Science, Problem 5[/i]
1. **Equation of the Sphere and Plane:**
   The sphere \( S \) is centered at the origin \( O \) with radius \(\sqrt{6}\). Its equation is:
   \[
   x^2 + y^2 + z^2 = 6
   \]
   The plane \(\alpha\) passes through the points \((4, 0, 0)\), \((0, 4, 0)\), and \((0, 0, 4)\). The equation of the plane can be found using these points. The general form of a plane equation is:
   \[
   ax + by + cz = d
   \]
   Substituting the points into the plane equation, we get:
   \[
   4a = d, \quad 4b = d, \quad 4c = d
   \]
   Since \(a = b = c\), we can set \(a = b = c = 1\) and \(d = 4\). Thus, the equation of the plane is:
   \[
   x + y + z = 4
   \]

2. **Intersection of the Sphere and Plane:**
   To find the common points of the sphere and the plane, we solve the system of equations:
   \[
   \begin{cases}
   x^2 + y^2 + z^2 = 6 \\
   x + y + z = 4
   \end{cases}
   \]
   We can use the second equation to express \(z\) in terms of \(x\) and \(y\):
   \[
   z = 4 - x - y
   \]
   Substituting \(z\) into the first equation:
   \[
   x^2 + y^2 + (4 - x - y)^2 = 6
   \]
   Expanding and simplifying:
   \[
   x^2 + y^2 + 16 - 8x - 8y + x^2 + y^2 + 2xy = 6
   \]
   \[
   2x^2 + 2y^2 + 2xy - 8x - 8y + 16 = 6
   \]
   \[
   2(x^2 + y^2 + xy - 4x - 4y + 8) = 12
   \]
   \[
   x^2 + y^2 + xy - 4x - 4y + 8 = 6
   \]
   \[
   x^2 + y^2 + xy - 4x - 4y + 2 = 0
   \]

3. **Using Vieta's Theorem:**
   Let \(x, y, z\) be the roots of the polynomial \(P(m) = m^3 + am^2 + bm + c\). By Vieta's formulas:
   \[
   -a = x + y + z = 4, \quad b = xy + yz + zx = 5, \quad -c = xyz
   \]
   Thus, the polynomial is:
   \[
   P(m) = m^3 - 4m^2 + 5m + c
   \]

4. **Finding the Range of \(xyz\):**
   To find the range of \(xyz\), we need to determine the values of \(c\) such that \(P(m)\) has three real roots. The derivative of \(P(m)\) is:
   \[
   P'(m) = 3m^2 - 8m + 5
   \]
   Solving \(P'(m) = 0\) for critical points:
   \[
   3m^2 - 8m + 5 = 0
   \]
   \[
   m = \frac{8 \pm \sqrt{64 - 60}}{6} = \frac{8 \pm 2}{6} = \frac{10}{6} = \frac{5}{3}, \quad \frac{6}{6} = 1
   \]
   Evaluating \(P(m)\) at these points:
   \[
   P(1) = 1^3 - 4 \cdot 1^2 + 5 \cdot 1 + c = 1 - 4 + 5 + c = 2 + c
   \]
   \[
   P\left(\frac{5}{3}\right) = \left(\frac{5}{3}\right)^3 - 4 \left(\frac{5}{3}\right)^2 + 5 \left(\frac{5}{3}\right) + c = \frac{125}{27} - \frac{100}{9} + \frac{25}{3} + c
   \]
   Simplifying:
   \[
   P\left(\frac{5}{3}\right) = \frac{125}{27} - \frac{300}{27} + \frac{225}{27} + c = \frac{50}{27} + c
   \]
   For \(P(m)\) to have three real roots, the values of \(c\) must satisfy:
   \[
   -P(1) \leq c \leq -P\left(\frac{5}{3}\right)
   \]
   \[
   -2 \leq c \leq -\frac{50}{27}
   \]
   Therefore, the range of \(xyz\) is:
   \[
   \frac{50}{27} \leq xyz \leq 2
   \]

The final answer is \(\boxed{\frac{50}{27} \leq xyz \leq 2}\)