If x,y,z are three real numbers such that ` x+ y +z =4 and x^(2) + y^(2) +z^(2) =6` ,then
(1) ` 2/3 le x,y,z le 2` (2) ` 0 le x,y,z le 2`
(3) ` 1 le x,y,z le 3` (4)` 2 le x,y,z le 3`

Answer

To solve the problem, we are given two equations involving three real numbers \( x, y, z \):

1. \( x + y + z = 4 \)
2. \( x^2 + y^2 + z^2 = 6 \)

We need to find the range of \( x, y, z \).

### Step 1: Express \( z \) in terms of \( x \) and \( y \)
From the first equation, we can express \( z \) as:
\[ z = 4 - x - y \]

### Step 2: Substitute \( z \) in the second equation
Now, substitute \( z \) into the second equation:
\[ x^2 + y^2 + (4 - x - y)^2 = 6 \]

### Step 3: Expand the equation
Expanding \( (4 - x - y)^2 \):
\[ (4 - x - y)^2 = 16 - 8x - 8y + x^2 + 2xy + y^2 \]
Thus, the equation becomes:
\[ x^2 + y^2 + 16 - 8x - 8y + x^2 + 2xy + y^2 = 6 \]
Combining like terms:
\[ 2x^2 + 2y^2 + 2xy - 8x - 8y + 16 = 6 \]
Simplifying gives:
\[ 2x^2 + 2y^2 + 2xy - 8x - 8y + 10 = 0 \]
Dividing the entire equation by 2:
\[ x^2 + y^2 + xy - 4x - 4y + 5 = 0 \]

### Step 4: Rearranging into a quadratic form
Rearranging gives:
\[ x^2 + (y - 4)x + (y^2 - 4y + 5) = 0 \]
This is a quadratic equation in \( x \).

### Step 5: Determine the discriminant
For \( x \) to be real, the discriminant must be non-negative:
\[ D = (y - 4)^2 - 4(y^2 - 4y + 5) \geq 0 \]
Calculating the discriminant:
\[ D = (y - 4)^2 - 4(y^2 - 4y + 5) = y^2 - 8y + 16 - 4y^2 + 16y - 20 \]
This simplifies to:
\[ -3y^2 + 8y - 4 \geq 0 \]

### Step 6: Solve the quadratic inequality
The roots of the equation \( -3y^2 + 8y - 4 = 0 \) can be found using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4 \cdot (-3) \cdot (-4)}}{2 \cdot (-3)} \]
Calculating the discriminant:
\[ 64 - 48 = 16 \]
Thus, the roots are:
\[ y = \frac{8 \pm 4}{-6} = \frac{12}{-6}, \frac{4}{-6} = -2, -\frac{2}{3} \]
The roots are \( \frac{2}{3} \) and \( 2 \).

### Step 7: Determine the intervals
The quadratic opens downwards (since the coefficient of \( y^2 \) is negative), so the inequality \( -3y^2 + 8y - 4 \geq 0 \) holds for:
\[ \frac{2}{3} \leq y \leq 2 \]

### Step 8: Conclude for \( x \) and \( z \)
By symmetry, since the equations are symmetric in \( x, y, z \), we have:
\[ \frac{2}{3} \leq x, y, z \leq 2 \]

### Final Answer
Thus, the correct option is:
(1) \( \frac{2}{3} \leq x, y, z \leq 2 \)