Use properties of determinants to evaluate :
`|{:(x+y,y+z,z+x),(z,x,y),(1,1,1):}|`

Answer

To evaluate the determinant

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

we will use properties of determinants step by step.

### Step 1: Add the second row to the first row.
We will perform the operation \( R_1 \rightarrow R_1 + R_2 \).

\[ D = \begin{vmatrix} (x+y) + z & (y+z) + x & (z+x) + y \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \]

This simplifies to:

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

### Step 2: Simplify the first row.
Notice that the first row can be rewritten as:

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

### Step 3: Factor out the common term from the first row.
We can factor out \( (x+y+z) \) from the first row:

\[ D = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 4: Identify identical rows.
Now, observe that the first and third rows are identical:

\[ D = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 5: Apply the property of determinants.
According to the property of determinants, if two rows (or columns) are identical, the value of the determinant is zero. Therefore:

\[ D = (x+y+z) \cdot 0 = 0 \]

### Conclusion
Thus, the value of the determinant is:

\[ \boxed{0} \]