Given is a prime number $p$ and natural $n$ such that $p \geq n \geq 3$. Set $A$ is made of sequences of lenght $n$ with elements from the set $\{0,1,2,...,p-1\}$ and have the following property:
  For arbitrary two sequence $(x_1,...,x_n)$ and $(y_1,...,y_n)$ from the set $A$ there exist three different numbers $k,l,m$ such that:

$x_k \not = y_k$, $x_l \not = y_l$, $x_m \not = y_m$.
Find the largest possible cardinality of $A$.
1. **Understanding the Problem:**
   We are given a prime number \( p \) and a natural number \( n \) such that \( p \geq n \geq 3 \). We need to find the largest possible cardinality of a set \( A \) of sequences of length \( n \) with elements from the set \( \{0, 1, 2, \ldots, p-1\} \) such that any two sequences in \( A \) differ in at least three positions.

2. **Upper Bound:**
   To find an upper bound, consider that any two sequences in \( A \) must differ in at least one of the first \( n-2 \) positions. This means that the number of sequences in \( A \) cannot exceed the number of sequences of length \( n-2 \) with elements from \( \{0, 1, 2, \ldots, p-1\} \). Therefore, the number of sequences in \( A \) is at most \( p^{n-2} \).

3. **Constructing Sequences:**
   We will construct \( p^{n-2} \) sequences that satisfy the given property. Consider sequences \( (x_1, x_2, \ldots, x_n) \) with elements from \( \{0, 1, 2, \ldots, p-1\} \) that satisfy the following conditions:
   \[
   x_{n-1} \equiv \sum_{i=1}^{n-2} x_i \pmod{p}
   \]
   \[
   x_n \equiv \sum_{i=1}^{n-2} i x_i \pmod{p}
   \]
   There are \( p^{n-2} \) such sequences because every sequence \( (x_1, x_2, \ldots, x_{n-2}) \) with elements from \( \{0, 1, 2, \ldots, p-1\} \) determines the values of \( x_{n-1} \) and \( x_n \).

4. **Verifying the Property:**
   We need to show that any two sequences \( (x_1, x_2, \ldots, x_n) \) and \( (y_1, y_2, \ldots, y_n) \) differ in at least three positions.

   - **Case 1:** The sequences \( (x_1, x_2, \ldots, x_{n-2}) \) and \( (y_1, y_2, \ldots, y_{n-2}) \) differ in at least three positions. There is nothing to prove in this case.

   - **Case 2:** The sequences \( (x_1, x_2, \ldots, x_{n-2}) \) and \( (y_1, y_2, \ldots, y_{n-2}) \) differ in exactly two positions, say \( i \) and \( j \) (where \( 1 \leq i < j \leq n-2 \)). Then \( x_i \neq y_i \) and \( x_j \neq y_j \). From the conditions, we have:
     \[
     x_{n-1} - y_{n-1} \equiv (x_i - y_i) + (x_j - y_j) \pmod{p}
     \]
     \[
     x_n - y_n \equiv i(x_i - y_i) + j(x_j - y_j) \pmod{p}
     \]
     Subtracting \( i \) times the first equation from the second, we get:
     \[
     (x_n - y_n) - i(x_{n-1} - y_{n-1}) \equiv (j - i)(x_j - y_j) \pmod{p}
     \]
     Since \( p \) is a prime number and \( x_j \neq y_j \), \( (j - i)(x_j - y_j) \) is not divisible by \( p \). Therefore, \( (x_n - y_n) - i(x_{n-1} - y_{n-1}) \) is not divisible by \( p \), implying that at least one of \( x_{n-1} \neq y_{n-1} \) or \( x_n \neq y_n \) holds. Thus, the sequences differ in at least three positions.

   - **Case 3:** The sequences \( (x_1, x_2, \ldots, x_{n-2}) \) and \( (y_1, y_2, \ldots, y_{n-2}) \) differ in exactly one position, say \( i \) (where \( 1 \leq i \leq n-2 \)). Then \( x_i \neq y_i \). From the conditions, we have:
     \[
     x_{n-1} - y_{n-1} \equiv x_i - y_i \pmod{p}
     \]
     \[
     x_n - y_n \equiv i(x_i - y_i) \pmod{p}
     \]
     Since \( x_i \neq y_i \), both \( x_{n-1} - y_{n-1} \) and \( x_n - y_n \) are not divisible by \( p \). Therefore, \( x_{n-1} \neq y_{n-1} \) and \( x_n \neq y_n \), implying that the sequences differ in at least three positions.

5. **Conclusion:**
   We have shown that the largest possible cardinality of \( A \) is \( p^{n-2} \).

The final answer is \( \boxed{p^{n-2}} \).