If a, b, and c are integers, then number of matrices `A=[(a,b,c),(b,c,a),(c,a,b)]` which are possible such that `A A^(T)=I` is _______.

Explanation

To solve the problem, we need to find the number of integer matrices \( A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \) such that \( A A^T = I \), where \( I \) is the identity matrix.

### Step-by-Step Solution:

1. **Write down the equation**:
We need to compute \( A A^T \) and set it equal to the identity matrix \( I \).

\[ A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \]
\[ A^T = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \]

2. **Calculate \( A A^T \)**:
We perform the multiplication of \( A \) and \( A^T \):

\[ A A^T = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \]

The elements of the resulting matrix are calculated as follows:

- First element (1,1):
\[ a^2 + b^2 + c^2 \]
- Second element (1,2):
\[ ab + bc + ca \]
- Third element (1,3):
\[ ac + ba + cb \]
- The same calculations apply for the other rows.

This results in the matrix:
\[ A A^T = \begin{pmatrix} a^2 + b^2 + c^2 & ab + bc + ca & ac + ba + cb \\ ab + bc + ca & b^2 + c^2 + a^2 & ba + cb + ac \\ ac + ba + cb & ba + cb + ac & c^2 + a^2 + b^2 \end{pmatrix} \]

3. **Set \( A A^T = I \)**:
We need:
\[ A A^T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

This gives us the following equations:
- \( a^2 + b^2 + c^2 = 1 \) (1)
- \( ab + bc + ca = 0 \) (2)

4. **Analyze the equations**:
From equation (1), since \( a, b, c \) are integers, the possible values for \( a^2, b^2, c^2 \) can only be \( 0 \) or \( 1 \). Therefore, the possible combinations of \( (a, b, c) \) are:
- One of \( a, b, c \) is \( \pm 1 \) and the others are \( 0 \).

5. **List the cases**:
- Case 1: \( a = \pm 1, b = 0, c = 0 \) → Matrices: \( \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix} \)
- Case 2: \( a = 0, b = \pm 1, c = 0 \) → Matrices: \( \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \)
- Case 3: \( a = 0, b = 0, c = \pm 1 \) → Matrices: \( \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{pmatrix} \)

6. **Count the valid matrices**:
Each case gives us 2 matrices, hence:
\[ \text{Total matrices} = 2 + 2 + 2 = 6 \]

### Final Answer:
The number of matrices \( A \) such that \( A A^T = I \) is **6**.