Find all polinomials $ P(x)$ with real coefficients, such that 

$ P(\sqrt {3}(a \minus{} b)) \plus{} P(\sqrt {3}(b \minus{} c)) \plus{} P(\sqrt {3}(c \minus{} a)) \equal{} P(2a \minus{} b \minus{} c) \plus{} P( \minus{} a \plus{} 2b \minus{} c) \plus{} P( \minus{} a \minus{} b \plus{} 2c)$
for any $ a$,$ b$ and $ c$ real numbers
1. Given the problem, we need to find all polynomials \( P(x) \) with real coefficients such that:
   \[
   P(\sqrt{3}(a - b)) + P(\sqrt{3}(b - c)) + P(\sqrt{3}(c - a)) = P(2a - b - c) + P(-a + 2b - c) + P(-a - b + 2c)
   \]
   for any real numbers \( a, b, \) and \( c \).

2. Denote \( x = a - c \), \( y = b - a \), and \( z = c - b \). Notice that \( x + y + z = 0 \).

3. The given equation can be rewritten using these new variables:
   \[
   P(\sqrt{3}x) + P(\sqrt{3}y) + P(\sqrt{3}z) = P(x - y) + P(y - z) + P(z - x)
   \]

4. Assume \( \deg P = n \). We compare the coefficients of the highest degree terms on both sides of the equation. This gives:
   \[
   (\sqrt{3})^n (x^n + y^n + z^n) = (x - y)^n + (y - z)^n + (z - x)^n
   \]

5. Let \( x = -1 \), \( y = 0 \), and \( z = 1 \). Then:
   \[
   (\sqrt{3})^n [(-1)^n + 0^n + 1^n] = (\sqrt{3})^n [(-1)^n + 1] = 2 (\sqrt{3})^n (-1)^n
   \]
   On the right-hand side:
   \[
   (x - y)^n + (y - z)^n + (z - x)^n = (-1 - 0)^n + (0 - 1)^n + (1 - (-1))^n = (-1)^n + (-1)^n + 2^n = 2(-1)^n + 2^n
   \]

6. Let \( x = 1 \), \( y = 0 \), and \( z = -1 \). Then:
   \[
   (\sqrt{3})^n [(1)^n + 0^n + (-1)^n] = (\sqrt{3})^n [1 + (-1)^n] = (\sqrt{3})^n (1 + (-1)^n)
   \]
   On the right-hand side:
   \[
   (x - y)^n + (y - z)^n + (z - x)^n = (1 - 0)^n + (0 - (-1))^n + (-1 - 1)^n = 1^n + 1^n + (-2)^n = 2 + (-2)^n
   \]

7. Combining the results from steps 5 and 6, we get:
   \[
   2 (\sqrt{3})^n (-1)^n = 2(-1)^n + 2^n
   \]
   and
   \[
   (\sqrt{3})^n (1 + (-1)^n) = 2 + (-2)^n
   \]

8. Simplifying these equations, we find:
   \[
   2^n + 2(-1)^n = 2 - 2^n
   \]
   \[
   2^n = 1 + (-1)^n
   \]

9. Solving \( 2^n = 1 + (-1)^n \):
   - If \( n = 0 \), then \( 2^0 = 1 + (-1)^0 \Rightarrow 1 = 1 + 1 \Rightarrow 1 = 2 \), which is false.
   - If \( n = 2 \), then \( 2^2 = 1 + (-1)^2 \Rightarrow 4 = 1 + 1 \Rightarrow 4 = 2 \), which is false.

10. Therefore, the only possible values for \( n \) are \( n = 0 \) and \( n = 2 \).

11. If \( n = 0 \), then \( P(x) = C \), a constant polynomial, which satisfies the given equation.

12. If \( n = 2 \), then \( P(x) = ax^2 + bx + c \), a quadratic polynomial, which also satisfies the given equation.

Conclusion:
The polynomials that satisfy the given equation are \( P(x) = C \) and \( P(x) = ax^2 + bx + c \).

The final answer is \( \boxed{ P(x) = C } \) and \( P(x) = ax^2 + bx + c \).