`|(""^(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))| =|(""^(x)C_r,""^(x+1)C_(r+1),""^(x+2)C_(r+2)),(""^(y)C_r,""^(y+1)C_(r+1),""^(y+2)C_(r+2)),(""^(z)C_r,""^(z+1)C_(r+1),""^(z+2)C_(r+2))|`

Answer

To solve the given determinant equality, we will analyze both sides of the equation step by step.

**Given:**
\[ | \binom{x}{r}, \binom{x}{r+1}, \binom{x}{r+2} | | \binom{y}{r}, \binom{y}{r+1}, \binom{y}{r+2} | | \binom{z}{r}, \binom{z}{r+1}, \binom{z}{r+2} | = | \binom{x}{r}, \binom{x+1}{r+1}, \binom{x+2}{r+2} | | \binom{y}{r}, \binom{y+1}{r+1}, \binom{y+2}{r+2} | | \binom{z}{r}, \binom{z+1}{r+1}, \binom{z+2}{r+2} | \]

### Step 1: Analyze the Left-Hand Side (LHS)
The LHS is:
\[ D_1 = \begin{vmatrix} \binom{x}{r} & \binom{x}{r+1} & \binom{x}{r+2} \\ \binom{y}{r} & \binom{y}{r+1} & \binom{y}{r+2} \\ \binom{z}{r} & \binom{z}{r+1} & \binom{z}{r+2} \end{vmatrix} \]

### Step 2: Apply Column Operations
We will perform column operations to simplify the determinant.

1. **Column 3:** Replace \(C_3\) with \(C_3 + C_2\):
\[ D_1 = \begin{vmatrix} \binom{x}{r} & \binom{x}{r+1} & \binom{x}{r+1} + \binom{x}{r+2} \\ \binom{y}{r} & \binom{y}{r+1} & \binom{y}{r+1} + \binom{y}{r+2} \\ \binom{z}{r} & \binom{z}{r+1} & \binom{z}{r+1} + \binom{z}{r+2} \end{vmatrix} \]

Using the identity \(\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}\), we can rewrite the last column:
\[ D_1 = \begin{vmatrix} \binom{x}{r} & \binom{x}{r+1} & \binom{x+1}{r+2} \\ \binom{y}{r} & \binom{y}{r+1} & \binom{y+1}{r+2} \\ \binom{z}{r} & \binom{z}{r+1} & \binom{z+1}{r+2} \end{vmatrix} \]

### Step 3: Further Column Operations
Next, we perform another column operation on \(C_2\):
2. **Column 2:** Replace \(C_2\) with \(C_2 + C_1\):
\[ D_1 = \begin{vmatrix} \binom{x}{r} & \binom{x}{r} + \binom{x}{r+1} & \binom{x+1}{r+2} \\ \binom{y}{r} & \binom{y}{r} + \binom{y}{r+1} & \binom{y+1}{r+2} \\ \binom{z}{r} & \binom{z}{r} + \binom{z}{r+1} & \binom{z+1}{r+2} \end{vmatrix} \]

Again, using the identity:
\[ D_1 = \begin{vmatrix} \binom{x}{r} & \binom{x+1}{r+1} & \binom{x+1}{r+2} \\ \binom{y}{r} & \binom{y+1}{r+1} & \binom{y+1}{r+2} \\ \binom{z}{r} & \binom{z+1}{r+1} & \binom{z+1}{r+2} \end{vmatrix} \]

### Step 4: Final Form of LHS
Now, we can see that the LHS has transformed into:
\[ D_1 = \begin{vmatrix} \binom{x}{r} & \binom{x+1}{r+1} & \binom{x+2}{r+2} \\ \binom{y}{r} & \binom{y+1}{r+1} & \binom{y+2}{r+2} \\ \binom{z}{r} & \binom{z+1}{r+1} & \binom{z+2}{r+2} \end{vmatrix} \]

### Step 5: Compare with Right-Hand Side (RHS)
The RHS is:
\[ D_2 = \begin{vmatrix} \binom{x}{r} & \binom{x+1}{r+1} & \binom{x+2}{r+2} \\ \binom{y}{r} & \binom{y+1}{r+1} & \binom{y+2}{r+2} \\ \binom{z}{r} & \binom{z+1}{r+1} & \binom{z+2}{r+2} \end{vmatrix} \]

### Conclusion
Since \(D_1\) and \(D_2\) are equal, we conclude that the given statement is **True**.

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