Find all the real $a,b,c$ such that the equality $|ax+by+cz| + |bx+cy+az| + |cx+ay+bz| = |x|+|y|+|z|$ is valid for all the real $x,y,z$.
1. **Substitute \( x = y = z = 1 \):**
   \[
   |a + b + c| + |b + c + a| + |c + a + b| = |1| + |1| + |1| = 3
   \]
   Since \( |a + b + c| \) appears three times, we have:
   \[
   3|a + b + c| = 3 \implies |a + b + c| = 1
   \]

2. **Substitute \( x = 1 \) and \( y = z = 0 \):**
   \[
   |a| + |b| + |c| = |1| + |0| + |0| = 1
   \]
   Thus, we have:
   \[
   |a| + |b| + |c| = 1
   \]

3. **Combine the results:**
   From the above, we have:
   \[
   |a + b + c| = 1 \quad \text{and} \quad |a| + |b| + |c| = 1
   \]
   This implies that \( a, b, \) and \( c \) must be of the same sign except the one(s) which is (are) equal to \( 0 \).

4. **Substitute \( (x, y, z) = (1, -1, 0) \):**
   \[
   |a - b| + |b - c| + |c - a| = |1| + |-1| + |0| = 2
   \]
   Using the triangle inequality:
   \[
   |a - b| + |b - c| + |c - a| \leq (|a| + |b|) + (|b| + |c|) + (|c| + |a|) = 2(|a| + |b| + |c|) = 2
   \]
   Since equality holds, we have:
   \[
   |a - b| = |a| + |b|, \quad |b - c| = |b| + |c|, \quad |c - a| = |c| + |a|
   \]
   This holds only if \( a \) and \( b \) are of opposite signs or at least one of them is equal to \( 0 \). Similarly for \( b \) and \( c \), and \( c \) and \( a \).

5. **Analyze the signs:**
   If none of \( a, b, \) and \( c \) is equal to \( 0 \), then:
   \[
   a > 0 \implies b < 0 \implies c > 0 \implies a < 0
   \]
   This is a contradiction. Therefore, at least two of \( a, b, \) and \( c \) must be \( 0 \).

6. **Possible cases:**
   The only possible cases are when two of \( a, b, \) and \( c \) are \( 0 \) and the third one is \( \pm 1 \). Thus, the solutions are:
   \[
   (a, b, c) = (\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1)
   \]

The final answer is \( \boxed{ (a, b, c) = (\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1) } \).