Find all triplets $ (x,y,z)$, that satisfy: 
$  \{\begin{array}{c}\ \ x^2 - 2x - 4z = 3\
y^2 - 2y - 2x = - 14 \
z^2 - 4y - 4z = - 18  \end{array}$
To find all triplets \((x, y, z)\) that satisfy the given system of equations:
\[
\begin{cases}
x^2 - 2x - 4z = 3 \\
y^2 - 2y - 2x = -14 \\
z^2 - 4y - 4z = -18
\end{cases}
\]

1. **Sum the equations:**
   \[
   x^2 - 2x - 4z + y^2 - 2y - 2x + z^2 - 4y - 4z = 3 - 14 - 18
   \]
   Simplify the right-hand side:
   \[
   3 - 14 - 18 = -29
   \]
   Combine like terms on the left-hand side:
   \[
   x^2 + y^2 + z^2 - 2x - 2y - 4y - 4z - 4z - 2x = -29
   \]
   Simplify:
   \[
   x^2 + y^2 + z^2 - 4x - 6y - 8z = -29
   \]

2. **Complete the square for each variable:**
   - For \(x\):
     \[
     x^2 - 4x = (x - 2)^2 - 4
     \]
   - For \(y\):
     \[
     y^2 - 6y = (y - 3)^2 - 9
     \]
   - For \(z\):
     \[
     z^2 - 8z = (z - 4)^2 - 16
     \]

3. **Substitute back into the equation:**
   \[
   (x - 2)^2 - 4 + (y - 3)^2 - 9 + (z - 4)^2 - 16 = -29
   \]
   Simplify:
   \[
   (x - 2)^2 + (y - 3)^2 + (z - 4)^2 - 29 = -29
   \]
   \[
   (x - 2)^2 + (y - 3)^2 + (z - 4)^2 = 0
   \]

4. **Solve for each variable:**
   Since the sum of squares is zero, each square must individually be zero:
   \[
   (x - 2)^2 = 0 \implies x = 2
   \]
   \[
   (y - 3)^2 = 0 \implies y = 3
   \]
   \[
   (z - 4)^2 = 0 \implies z = 4
   \]

Thus, the only solution to the system of equations is:
\[
(x, y, z) = (2, 3, 4)
\]

The final answer is \( \boxed{ (2, 3, 4) } \)