`|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z)C_(r+2)):}|` is equal to

Explanation

To solve the determinant

\[ D = \begin{vmatrix} \binom{x}{r} & \binom{x}{r+1} & \binom{x}{r+2} \\ \binom{y}{r} & \binom{y}{r+1} & \binom{y}{r+2} \\ \binom{z}{r} & \binom{z}{r+1} & \binom{z}{r+2} \end{vmatrix} \]

we will perform a series of column operations and apply the known identity for binomial coefficients.

### Step 1: Perform Column Operations
Let's perform the operation \( C_2 \to C_2 + C_1 \) and \( C_3 \to C_3 + C_1 \).

After this operation, the determinant becomes:

\[ D = \begin{vmatrix} \binom{x}{r} & \binom{x}{r} + \binom{x}{r+1} & \binom{x}{r} + \binom{x}{r+2} \\ \binom{y}{r} & \binom{y}{r} + \binom{y}{r+1} & \binom{y}{r} + \binom{y}{r+2} \\ \binom{z}{r} & \binom{z}{r} + \binom{z}{r+1} & \binom{z}{r} + \binom{z}{r+2} \end{vmatrix} \]

### Step 2: Apply the Binomial Coefficient Identity
Using the identity \( \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} \), we can simplify the second and third columns:

- For \( C_2 \):
\[ \binom{x}{r} + \binom{x}{r+1} = \binom{x+1}{r+1} \]
- For \( C_3 \):
\[ \binom{x}{r} + \binom{x}{r+2} = \binom{x+1}{r+2} \]

So, the determinant now looks like:

\[ D = \begin{vmatrix} \binom{x}{r} & \binom{x+1}{r+1} & \binom{x+1}{r+2} \\ \binom{y}{r} & \binom{y+1}{r+1} & \binom{y+1}{r+2} \\ \binom{z}{r} & \binom{z+1}{r+1} & \binom{z+1}{r+2} \end{vmatrix} \]

### Step 3: Repeat the Process
Now, we can repeat the same process for the second column. Perform \( C_3 \to C_3 + C_2 \):

After this operation, we have:

\[ D = \begin{vmatrix} \binom{x}{r} & \binom{x+1}{r+1} & \binom{x+1}{r+1} + \binom{x+1}{r+2} \\ \binom{y}{r} & \binom{y+1}{r+1} & \binom{y+1}{r+1} + \binom{y+1}{r+2} \\ \binom{z}{r} & \binom{z+1}{r+1} & \binom{z+1}{r+1} + \binom{z+1}{r+2} \end{vmatrix} \]

Using the same identity again, we can simplify the third column:

\[ \binom{a}{b} + \binom{a}{b+1} = \binom{a+1}{b+1} \]

### Final Result
After applying the identity, we can conclude that the determinant simplifies to:

\[ D = \begin{vmatrix} \binom{x}{r} & \binom{x+1}{r+1} & \binom{x+2}{r+2} \\ \binom{y}{r} & \binom{y+1}{r+1} & \binom{y+2}{r+2} \\ \binom{z}{r} & \binom{z+1}{r+1} & \binom{z+2}{r+2} \end{vmatrix} \]

### Conclusion
Thus, the determinant is equal to one of the options provided in the question. After checking through the options, we find that all four options are indeed correct.