Evaluate `|{:(.^(x)C_(1),,.^(x)C_(2),,.^(x)C_(3)),(.^(y)C_(1),,.^(y)C_(2),,.^(y)C_(3)),(.^(x)C_(1),,.^(z)C_(2),,.^(z)C_(3)):}|`

Explanation

To evaluate the determinant

\[ D = \begin{vmatrix} {x \choose 1} & {x \choose 2} & {x \choose 3} \\ {y \choose 1} & {y \choose 2} & {y \choose 3} \\ {z \choose 1} & {z \choose 2} & {z \choose 3} \end{vmatrix} \]

we will follow these steps:

### Step 1: Write the Binomial Coefficients
First, we will express the binomial coefficients in terms of \(x\), \(y\), and \(z\):

\[ {x \choose 1} = x, \quad {x \choose 2} = \frac{x(x-1)}{2}, \quad {x \choose 3} = \frac{x(x-1)(x-2)}{6} \]

Similarly, for \(y\) and \(z\):

\[ {y \choose 1} = y, \quad {y \choose 2} = \frac{y(y-1)}{2}, \quad {y \choose 3} = \frac{y(y-1)(y-2)}{6} \]

\[ {z \choose 1} = z, \quad {z \choose 2} = \frac{z(z-1)}{2}, \quad {z \choose 3} = \frac{z(z-1)(z-2)}{6} \]

Thus, we can rewrite the determinant as:

\[ D = \begin{vmatrix} x & \frac{x(x-1)}{2} & \frac{x(x-1)(x-2)}{6} \\ y & \frac{y(y-1)}{2} & \frac{y(y-1)(y-2)}{6} \\ z & \frac{z(z-1)}{2} & \frac{z(z-1)(z-2)}{6} \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Next, we can factor out common terms from each row:

- From the first row, factor out \(x\).
- From the second row, factor out \(\frac{y}{2}\).
- From the third row, factor out \(\frac{z}{6}\).

This gives us:

\[ D = x \cdot \frac{y}{2} \cdot \frac{z}{6} \cdot \begin{vmatrix} 1 & (x-1) & \frac{(x-1)(x-2)}{3} \\ 1 & (y-1) & \frac{(y-1)(y-2)}{3} \\ 1 & (z-1) & \frac{(z-1)(z-2)}{3} \end{vmatrix} \]

### Step 3: Simplify the Determinant
Now, we can simplify the determinant:

\[ D = \frac{xyz}{12} \cdot \begin{vmatrix} 1 & (x-1) & \frac{(x-1)(x-2)}{3} \\ 1 & (y-1) & \frac{(y-1)(y-2)}{3} \\ 1 & (z-1) & \frac{(z-1)(z-2)}{3} \end{vmatrix} \]

### Step 4: Perform Column Operations
Next, we can perform column operations to simplify the determinant further. We can replace the third column with the third column minus the second column multiplied by \(\frac{(x-1)}{3}\):

\[ D = \frac{xyz}{12} \cdot \begin{vmatrix} 1 & (x-1) & 0 \\ 1 & (y-1) & 0 \\ 1 & (z-1) & 0 \end{vmatrix} \]

### Step 5: Calculate the Determinant
The determinant of a matrix with two identical columns is zero. Therefore, we can conclude that:

\[ D = 0 \]

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

\[ D = 0 \]