`if A=[{:(3,2,1),(1,2,3),(3,-6,1):}],B=[{:(1,4,0),(2,-3,0),(1,2,0):}]`
`and C=[{:(1,2,3),(3,2,1),(8,7,9):}],`then find AB-AC.

Explanation

To solve the problem, we need to find \( AB - AC \) where the matrices \( A \), \( B \), and \( C \) are given as follows:

\[ A = \begin{pmatrix} 3 & 2 & 1 \\ 1 & 2 & 3 \\ 3 & -6 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 4 & 0 \\ 2 & -3 & 0 \\ 1 & 2 & 0 \end{pmatrix}, \quad C = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 8 & 7 & 9 \end{pmatrix} \]

### Step 1: Calculate \( AB \)

To find \( AB \), we perform matrix multiplication:

\[ AB = A \cdot B \]

Calculating each element of the resulting matrix:

- First row:
- \( (3 \cdot 1 + 2 \cdot 2 + 1 \cdot 1) = 3 + 4 + 1 = 8 \)
- \( (3 \cdot 4 + 2 \cdot -3 + 1 \cdot 2) = 12 - 6 + 2 = 8 \)
- \( (3 \cdot 0 + 2 \cdot 0 + 1 \cdot 0) = 0 \)

- Second row:
- \( (1 \cdot 1 + 2 \cdot 2 + 3 \cdot 1) = 1 + 4 + 3 = 8 \)
- \( (1 \cdot 4 + 2 \cdot -3 + 3 \cdot 2) = 4 - 6 + 6 = 4 \)
- \( (1 \cdot 0 + 2 \cdot 0 + 3 \cdot 0) = 0 \)

- Third row:
- \( (3 \cdot 1 + -6 \cdot 2 + 1 \cdot 1) = 3 - 12 + 1 = -8 \)
- \( (3 \cdot 4 + -6 \cdot -3 + 1 \cdot 2) = 12 + 18 + 2 = 32 \)
- \( (3 \cdot 0 + -6 \cdot 0 + 1 \cdot 0) = 0 \)

Thus, we have:

\[ AB = \begin{pmatrix} 8 & 8 & 0 \\ 8 & 4 & 0 \\ -8 & 32 & 0 \end{pmatrix} \]

### Step 2: Calculate \( AC \)

Next, we calculate \( AC \):

\[ AC = A \cdot C \]

Calculating each element of the resulting matrix:

- First row:
- \( (3 \cdot 1 + 2 \cdot 3 + 1 \cdot 8) = 3 + 6 + 8 = 17 \)
- \( (3 \cdot 2 + 2 \cdot 2 + 1 \cdot 7) = 6 + 4 + 7 = 17 \)
- \( (3 \cdot 3 + 2 \cdot 1 + 1 \cdot 9) = 9 + 2 + 9 = 20 \)

- Second row:
- \( (1 \cdot 1 + 2 \cdot 3 + 3 \cdot 8) = 1 + 6 + 24 = 31 \)
- \( (1 \cdot 2 + 2 \cdot 2 + 3 \cdot 7) = 2 + 4 + 21 = 27 \)
- \( (1 \cdot 3 + 2 \cdot 1 + 3 \cdot 9) = 3 + 2 + 27 = 32 \)

- Third row:
- \( (3 \cdot 1 + -6 \cdot 3 + 1 \cdot 8) = 3 - 18 + 8 = -7 \)
- \( (3 \cdot 2 + -6 \cdot 2 + 1 \cdot 7) = 6 - 12 + 7 = 1 \)
- \( (3 \cdot 3 + -6 \cdot 1 + 1 \cdot 9) = 9 - 6 + 9 = 12 \)

Thus, we have:

\[ AC = \begin{pmatrix} 17 & 17 & 20 \\ 31 & 27 & 32 \\ -7 & 1 & 12 \end{pmatrix} \]

### Step 3: Calculate \( AB - AC \)

Now, we subtract \( AC \) from \( AB \):

\[ AB - AC = \begin{pmatrix} 8 & 8 & 0 \\ 8 & 4 & 0 \\ -8 & 32 & 0 \end{pmatrix} - \begin{pmatrix} 17 & 17 & 20 \\ 31 & 27 & 32 \\ -7 & 1 & 12 \end{pmatrix} \]

Calculating each element:

- First row:
- \( 8 - 17 = -9 \)
- \( 8 - 17 = -9 \)
- \( 0 - 20 = -20 \)

- Second row:
- \( 8 - 31 = -23 \)
- \( 4 - 27 = -23 \)
- \( 0 - 32 = -32 \)

- Third row:
- \( -8 - (-7) = -1 \)
- \( 32 - 1 = 31 \)
- \( 0 - 12 = -12 \)

Thus, we have:

\[ AB - AC = \begin{pmatrix} -9 & -9 & -20 \\ -23 & -23 & -32 \\ -1 & 31 & -12 \end{pmatrix} \]

### Final Answer

The final result is:

\[ AB - AC = \begin{pmatrix} -9 & -9 & -20 \\ -23 & -23 & -32 \\ -1 & 31 & -12 \end{pmatrix} \]