Find all polynomials $p(x)$ with real coeffcients such that
\[p(a + b - 2c) + p(b + c - 2a) + p(c + a - 2b) = 3p(a - b) + 3p(b - c) + 3p(c - a)\]
for all $a, b, c\in\mathbb{R}$.

[i](2nd Benelux Mathematical Olympiad 2010, Problem 2)[/i]
1. **Substitution and Simplification:**
   We start by substituting \( x = a - b \) and \( y = b - c \) into the given equation:
   \[
   p(a + b - 2c) + p(b + c - 2a) + p(c + a - 2b) = 3p(a - b) + 3p(b - c) + 3p(c - a)
   \]
   This transforms the equation into:
   \[
   p(x - y) + p(-2x - y) + p(x + 2y) = 3p(x) + 3p(y) + 3p(-x - y)
   \]

2. **Special Case \( x = y = 0 \):**
   Setting \( x = 0 \) and \( y = 0 \) in the transformed equation, we get:
   \[
   p(0) + p(0) + p(0) = 3p(0) + 3p(0) + 3p(0)
   \]
   Simplifying, we find:
   \[
   3p(0) = 9p(0) \implies p(0) = 0
   \]

3. **Special Case \( y = 0 \):**
   Setting \( y = 0 \) in the transformed equation, we get:
   \[
   p(x) + p(-2x) + p(x) = 3p(x) + 3p(-x)
   \]
   Simplifying, we find:
   \[
   2p(x) + p(-2x) = 3p(x) + 3p(-x) \implies p(-2x) = p(x) + 3p(-x)
   \]

4. **Assuming Polynomial Form:**
   Assume \( p(x) \) is a polynomial of degree \( n \):
   \[
   p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
   \]
   Since \( p(0) = 0 \), we have \( a_0 = 0 \). Thus:
   \[
   p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x
   \]

5. **Analyzing the Equation \( p(-2x) = p(x) + 3p(-x) \):**
   Substitute \( p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x \) into the equation:
   \[
   a_n (-2x)^n + a_{n-1} (-2x)^{n-1} + \cdots + a_1 (-2x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + 3(a_n (-x)^n + a_{n-1} (-x)^{n-1} + \cdots + a_1 (-x))
   \]
   Simplifying, we get:
   \[
   a_n (-2)^n x^n + a_{n-1} (-2)^{n-1} x^{n-1} + \cdots + a_1 (-2) x = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + 3(a_n (-1)^n x^n + a_{n-1} (-1)^{n-1} x^{n-1} + \cdots + a_1 (-1) x)
   \]

6. **Equating Coefficients:**
   For each power of \( x \), we equate the coefficients:
   \[
   a_k (-2)^k = a_k + 3a_k (-1)^k
   \]
   This simplifies to:
   \[
   a_k \left[ (-2)^k - 1 - 3(-1)^k \right] = 0
   \]
   This gives us two cases:
   \[
   \begin{cases}
   a_k (2^k - 4) = 0 & \text{if } k \text{ is even} \\
   a_k (2^k - 2) = 0 & \text{if } k \text{ is odd}
   \end{cases}
   \]

7. **Solving for Coefficients:**
   For \( k \geq 3 \), both equations yield \( a_k = 0 \). Therefore, the polynomial \( p(x) \) must be of degree at most 2:
   \[
   p(x) = \lambda x^2 + \mu x
   \]
   where \( \lambda \) and \( \mu \) are real constants.

8. **Verification:**
   Substituting \( p(x) = \lambda x^2 + \mu x \) back into the original equation, we find that it satisfies the equation for any \( \lambda, \mu \in \mathbb{R} \).

The final answer is \( \boxed{ p(x) = \lambda x^2 + \mu x } \), where \( \lambda, \mu \in \mathbb{R} \).