The value of `|(.^(10)C_(4).^(10)C_(5).^(11)C_(m)),(.^(11)C_(6).^(11)C_(7).^(12)C_(m+2)),(.^(12)C_(8).^(12)C_(9).^(13)C_(m+4))|` is equal to zero when `m` is

Options

A. `6`
B. `4`
C. `5`
D. `7`

Explanation

To solve the problem, we need to evaluate the determinant given by:

\[ D = \begin{vmatrix} \binom{10}{4} & \binom{10}{5} & \binom{11}{m} \\ \binom{11}{6} & \binom{11}{7} & \binom{12}{m+2} \\ \binom{12}{8} & \binom{12}{9} & \binom{13}{m+4} \end{vmatrix} \]

We want to find the value of \( m \) for which this determinant equals zero.

### Step 1: Apply Column Operations
We can use the property of determinants that allows us to add columns. Specifically, we can add the second column to the first column:

\[ C_1 \rightarrow C_1 + C_2 \]

This gives us:

\[ D = \begin{vmatrix} \binom{10}{4} + \binom{10}{5} & \binom{10}{5} & \binom{11}{m} \\ \binom{11}{6} + \binom{11}{7} & \binom{11}{7} & \binom{12}{m+2} \\ \binom{12}{8} + \binom{12}{9} & \binom{12}{9} & \binom{13}{m+4} \end{vmatrix} \]

### Step 2: Simplify Using Combinatorial Identities
Using the identity \( \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \):

1. For the first row:
\[ \binom{10}{4} + \binom{10}{5} = \binom{11}{5} \]

2. For the second row:
\[ \binom{11}{6} + \binom{11}{7} = \binom{12}{7} \]

3. For the third row:
\[ \binom{12}{8} + \binom{12}{9} = \binom{13}{9} \]

Thus, we can rewrite the determinant as:

\[ D = \begin{vmatrix} \binom{11}{5} & \binom{10}{5} & \binom{11}{m} \\ \binom{12}{7} & \binom{11}{7} & \binom{12}{m+2} \\ \binom{13}{9} & \binom{12}{9} & \binom{13}{m+4} \end{vmatrix} \]

### Step 3: Analyze the Determinant
The determinant \( D \) will be zero if any two rows or columns are identical.

### Step 4: Set Conditions for Equality
To find \( m \), we need to set conditions for the columns to be equal.

1. Set \( m = 5 \):
- Then \( \binom{11}{m} = \binom{11}{5} \)
- \( \binom{12}{m+2} = \binom{12}{7} \)
- \( \binom{13}{m+4} = \binom{13}{9} \)

This gives us:

\[ D = \begin{vmatrix} \binom{11}{5} & \binom{10}{5} & \binom{11}{5} \\ \binom{12}{7} & \binom{11}{7} & \binom{12}{7} \\ \binom{13}{9} & \binom{12}{9} & \binom{13}{9} \end{vmatrix} \]

### Step 5: Conclusion
Since the first and third columns are identical, the determinant \( D \) equals zero.

Thus, the value of \( m \) for which the determinant is zero is:

\[ \boxed{5} \]