the value of the determinant
`|{:(.^(n)C_(r-1),,.^(n)C_(r),,(r+1)^(n+2)C_(r+1)),(.^(n)C_(r),,.^(n)C_(r+1),,(r+2)^(n+2)C_(r+2)),(.^(n)C_(r+1),,.^(n)C_(r+2),,(r+3)^(n+2)C_(r+3)):}|` is

Options

A. `n^(2)+n-1)`
B. `0`
C. `.^(n+3)C_(r+3)`
D. `.^(n)C_(r-1)+^(n)C_(r)+^(n)C_(r+1)`

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} {n \choose r-1} & {n \choose r} & (r+1) {n+2 \choose r+1} \\ {n \choose r} & {n \choose r+1} & (r+2) {n+2 \choose r+2} \\ {n \choose r+1} & {n \choose r+2} & (r+3) {n+2 \choose r+3} \end{vmatrix} \]

we will follow these steps:

### Step 1: Simplify the third column
The elements of the third column can be expressed in terms of binomial coefficients. We can factor out the common terms in the third column.

- The first element is \((r+1) {n+2 \choose r+1}\)
- The second element is \((r+2) {n+2 \choose r+2}\)
- The third element is \((r+3) {n+2 \choose r+3}\)

Factoring out \({n+2}\) from the third column gives us:

\[ D = {n+2} \begin{vmatrix} {n \choose r-1} & {n \choose r} & {n+1 \choose r+1} \\ {n \choose r} & {n \choose r+1} & {n+1 \choose r+2} \\ {n \choose r+1} & {n \choose r+2} & {n+1 \choose r+3} \end{vmatrix} \]

### Step 2: Apply the identity for binomial coefficients
Using the identity \( {n \choose r} + {n \choose r+1} = {n+1 \choose r+1} \), we can manipulate the determinant. We will add the first column to the second column:

\[ D = {n+2} \begin{vmatrix} {n \choose r-1} & {n+1 \choose r} & {n+1 \choose r+1} \\ {n \choose r} & {n+1 \choose r+1} & {n+1 \choose r+2} \\ {n \choose r+1} & {n+1 \choose r+2} & {n+1 \choose r+3} \end{vmatrix} \]

### Step 3: Identify identical columns
Notice that the first and third columns are identical:

\[ \begin{vmatrix} {n \choose r-1} & {n+1 \choose r} & {n+1 \choose r} \\ {n \choose r} & {n+1 \choose r+1} & {n+1 \choose r+1} \\ {n \choose r+1} & {n+1 \choose r+2} & {n+1 \choose r+2} \end{vmatrix} \]

### Step 4: Conclude the determinant value
Since two columns of the determinant are identical, the value of the determinant is zero:

\[ D = 0 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{0} \]