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 $x^2 + y^2 + z^2$?
1. We start with the given equations:
   \[
   x^2 + 6y = -17
   \]
   \[
   y^2 + 4z = 1
   \]
   \[
   z^2 + 2x = 2
   \]

2. Add the three equations together:
   \[
   x^2 + 6y + y^2 + 4z + z^2 + 2x = -17 + 1 + 2
   \]
   Simplifying the right-hand side:
   \[
   x^2 + y^2 + z^2 + 2x + 6y + 4z = -14
   \]

3. Complete the square for each variable:
   - For \(x\):
     \[
     x^2 + 2x = (x+1)^2 - 1
     \]
   - For \(y\):
     \[
     y^2 + 6y = (y+3)^2 - 9
     \]
   - For \(z\):
     \[
     z^2 + 4z = (z+2)^2 - 4
     \]

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

5. Combine constants on the left-hand side:
   \[
   (x+1)^2 + (y+3)^2 + (z+2)^2 - 14 = -14
   \]

6. Simplify the equation:
   \[
   (x+1)^2 + (y+3)^2 + (z+2)^2 = 0
   \]

7. Since each term in the sum of squares is nonnegative, the only way their sum can be zero is if each term is zero:
   \[
   (x+1)^2 = 0 \implies x = -1
   \]
   \[
   (y+3)^2 = 0 \implies y = -3
   \]
   \[
   (z+2)^2 = 0 \implies z = -2
   \]

8. Now, calculate \(x^2 + y^2 + z^2\):
   \[
   x^2 + y^2 + z^2 = (-1)^2 + (-3)^2 + (-2)^2 = 1 + 9 + 4 = 14
   \]

The final answer is \(\boxed{14}\).