Three real numbers x, y, z are such that `x^(2) + 6y =-17, y^(2) + 4z=1` and `z^(2) + 2x=2`. What is the value of `x^(2) + y^(2) + z^(2)`.

Answer

To solve the problem, we have three equations involving the real numbers \(x\), \(y\), and \(z\):

1. \(x^2 + 6y = -17\) (Equation 1)
2. \(y^2 + 4z = 1\) (Equation 2)
3. \(z^2 + 2x = 2\) (Equation 3)

We need to find the value of \(x^2 + y^2 + z^2\).

### Step 1: Rearranging the equations
First, we can rearrange each equation to express \(y\), \(z\), and \(x\) in terms of the other variables.

From Equation 1:
\[ 6y = -17 - x^2 \implies y = \frac{-17 - x^2}{6} \]

From Equation 2:
\[ 4z = 1 - y^2 \implies z = \frac{1 - y^2}{4} \]

From Equation 3:
\[ 2x = 2 - z^2 \implies x = \frac{2 - z^2}{2} \]

### Step 2: Adding the equations
Next, we add all three equations together:
\[ (x^2 + 6y) + (y^2 + 4z) + (z^2 + 2x) = -17 + 1 + 2 \]
This simplifies to:
\[ x^2 + y^2 + z^2 + 6y + 4z + 2x = -14 \]

### Step 3: Grouping terms
Now we group the terms:
\[ x^2 + 2x + y^2 + 6y + z^2 + 4z = -14 \]

### Step 4: Completing the square
We will complete the square for each variable:

1. For \(x^2 + 2x\):
\[ x^2 + 2x = (x + 1)^2 - 1 \]

2. For \(y^2 + 6y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

3. For \(z^2 + 4z\):
\[ z^2 + 4z = (z + 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ ((x + 1)^2 - 1) + ((y + 3)^2 - 9) + ((z + 2)^2 - 4) = -14 \]

Combining the constants:
\[ (x + 1)^2 + (y + 3)^2 + (z + 2)^2 - 14 = -14 \]

### Step 5: Simplifying the equation
This simplifies to:
\[ (x + 1)^2 + (y + 3)^2 + (z + 2)^2 = 0 \]

### Step 6: Finding the values of \(x\), \(y\), and \(z\)
Since the sum of squares is zero, each square must be zero:
\[ x + 1 = 0 \implies x = -1 \]
\[ y + 3 = 0 \implies y = -3 \]
\[ z + 2 = 0 \implies z = -2 \]

### Step 7: Calculating \(x^2 + y^2 + z^2\)
Now we can find \(x^2 + y^2 + z^2\):
\[ x^2 + y^2 + z^2 = (-1)^2 + (-3)^2 + (-2)^2 = 1 + 9 + 4 = 14 \]

### Final Answer
Thus, the value of \(x^2 + y^2 + z^2\) is:
\[ \boxed{14} \]