If `|{:(.^(9)C_(4),.^(9)C_(5),.^(10)C_(r)),(.^(10)C_(6 ),.^(10)C_(7),.^(11)C_(r+2)),(.^(11)C_(8),.^(11)C_(9),.^(12)C_(r+4)):}|=0` , then the value of `r` is equal to

Options

A. `3`
B. `4`
C. `5`
D. `6`

Explanation

To solve the determinant equation given by

\[ \begin{vmatrix} {9C_4} & {9C_5} & {10C_r} \\ {10C_6} & {10C_7} & {11C_{r+2}} \\ {11C_8} & {11C_9} & {12C_{r+4}} \end{vmatrix} = 0, \]

we will follow these steps:

### Step 1: Write down the determinant
We start with the determinant as given:

\[ D = \begin{vmatrix} {9C_4} & {9C_5} & {10C_r} \\ {10C_6} & {10C_7} & {11C_{r+2}} \\ {11C_8} & {11C_9} & {12C_{r+4}} \end{vmatrix} \]

### Step 2: Apply the column operation
We can simplify the determinant by applying the column operation \(C_2 \rightarrow C_2 + C_3\):

\[ D = \begin{vmatrix} {9C_4} & {9C_5 + 10C_r} & {10C_r} \\ {10C_6} & {10C_7 + 11C_{r+2}} & {11C_{r+2}} \\ {11C_8} & {11C_9 + 12C_{r+4}} & {12C_{r+4}} \end{vmatrix} \]

### Step 3: Analyze the determinant
For the determinant to equal zero, at least two columns must be linearly dependent. This can happen if:

\[ 9C_5 + 10C_r = 10C_r \quad \text{(1st and 2nd column)} \]
or
\[ 10C_7 + 11C_{r+2} = 11C_{r+2} \quad \text{(2nd and 3rd column)} \]
or
\[ 11C_9 + 12C_{r+4} = 12C_{r+4} \quad \text{(3rd and 4th column)} \]

### Step 4: Set up the equation
From the first column operation, we have:

\[ 9C_5 = 10C_r - 10C_r \implies 9C_5 = 0 \implies C_r = C_{r+2} = C_{r+4} \]

### Step 5: Solve for \(r\)
We can analyze the values of \(r\) based on the properties of combinations. The values of \(C\) can be equal if:

- \(r = 5\)
- \(r + 2 = 5\) or \(r + 4 = 5\)

Thus, we can conclude that:

\[ r = 5 \]

### Conclusion
The value of \(r\) that satisfies the determinant condition is:

\[ \boxed{5} \]