Let `A=[(0,6,7),(-6,0,8),(7,-8,0)],B=[(0,1,1),(1,0,2),(1,2,0)],C=[(2),(-2),(3)]`.
Calculate AC, BC and (A+B)C.
Also show that (A+B)C=AC+BC.

Answer

To solve the problem, we need to calculate \( AC \), \( BC \), and \( (A + B)C \), and then verify that \( (A + B)C = AC + BC \).

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

Given:
\[ A = \begin{pmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{pmatrix}, \quad C = \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} \]

To find \( AC \):
\[ AC = A \cdot C = \begin{pmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} \]

Calculating each element:
1. First row: \( 0 \cdot 2 + 6 \cdot (-2) + 7 \cdot 3 = 0 - 12 + 21 = 9 \)
2. Second row: \( -6 \cdot 2 + 0 \cdot (-2) + 8 \cdot 3 = -12 + 0 + 24 = 12 \)
3. Third row: \( 7 \cdot 2 + (-8) \cdot (-2) + 0 \cdot 3 = 14 + 16 + 0 = 30 \)

Thus,
\[ AC = \begin{pmatrix} 9 \\ 12 \\ 30 \end{pmatrix} \]

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

Given:
\[ B = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{pmatrix} \]

To find \( BC \):
\[ BC = B \cdot C = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} \]

Calculating each element:
1. First row: \( 0 \cdot 2 + 1 \cdot (-2) + 1 \cdot 3 = 0 - 2 + 3 = 1 \)
2. Second row: \( 1 \cdot 2 + 0 \cdot (-2) + 2 \cdot 3 = 2 + 0 + 6 = 8 \)
3. Third row: \( 1 \cdot 2 + 2 \cdot (-2) + 0 \cdot 3 = 2 - 4 + 0 = -2 \)

Thus,
\[ BC = \begin{pmatrix} 1 \\ 8 \\ -2 \end{pmatrix} \]

### Step 3: Calculate \( A + B \)

To find \( A + B \):
\[ A + B = \begin{pmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{pmatrix} \]

Calculating each element:
1. First row: \( 0 + 0, 6 + 1, 7 + 1 = 0, 7, 8 \)
2. Second row: \( -6 + 1, 0 + 0, 8 + 2 = -5, 0, 10 \)
3. Third row: \( 7 + 1, -8 + 2, 0 + 0 = 8, -6, 0 \)

Thus,
\[ A + B = \begin{pmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0 \end{pmatrix} \]

### Step 4: Calculate \( (A + B)C \)

Now, we compute \( (A + B)C \):
\[ (A + B)C = \begin{pmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} \]

Calculating each element:
1. First row: \( 0 \cdot 2 + 7 \cdot (-2) + 8 \cdot 3 = 0 - 14 + 24 = 10 \)
2. Second row: \( -5 \cdot 2 + 0 \cdot (-2) + 10 \cdot 3 = -10 + 0 + 30 = 20 \)
3. Third row: \( 8 \cdot 2 + (-6) \cdot (-2) + 0 \cdot 3 = 16 + 12 + 0 = 28 \)

Thus,
\[ (A + B)C = \begin{pmatrix} 10 \\ 20 \\ 28 \end{pmatrix} \]

### Step 5: Verify \( (A + B)C = AC + BC \)

Now we check if \( (A + B)C = AC + BC \):
\[ AC + BC = \begin{pmatrix} 9 \\ 12 \\ 30 \end{pmatrix} + \begin{pmatrix} 1 \\ 8 \\ -2 \end{pmatrix} \]

Calculating:
1. First row: \( 9 + 1 = 10 \)
2. Second row: \( 12 + 8 = 20 \)
3. Third row: \( 30 - 2 = 28 \)

Thus,
\[ AC + BC = \begin{pmatrix} 10 \\ 20 \\ 28 \end{pmatrix} \]

Since \( (A + B)C = AC + BC \), we have verified the equation.

### Final Answers:
- \( AC = \begin{pmatrix} 9 \\ 12 \\ 30 \end{pmatrix} \)
- \( BC = \begin{pmatrix} 1 \\ 8 \\ -2 \end{pmatrix} \)
- \( (A + B)C = \begin{pmatrix} 10 \\ 20 \\ 28 \end{pmatrix} \)