evaluate: `|(3x,-x+y,-x+z),(x-y,3y,z-y),(x-z,y-z,3z)|`

Explanation

To evaluate the determinant \( D = \begin{vmatrix} 3x & -x+y & -x+z \\ x-y & 3y & z-y \\ x-z & y-z & 3z \end{vmatrix} \), we will follow a systematic approach.

### Step 1: Apply Column Transformation
We will transform the first column \( C_1 \) by adding the second column \( C_2 \) and the third column \( C_3 \) to it. This gives us:
\[ C_1 \rightarrow C_1 + C_2 + C_3 \]
After this transformation, the determinant becomes:
\[ D = \begin{vmatrix} 3x + (-x+y) + (-x+z) & -x+y & -x+z \\ (x-y) + 3y + (z-y) & 3y & z-y \\ (x-z) + (y-z) + 3z & y-z & 3z \end{vmatrix} \]
Calculating the first column:
- First row: \( 3x - x + y - x + z = x + y + z \)
- Second row: \( x - y + 3y + z - y = x + 2y + z \)
- Third row: \( x - z + y - z + 3z = x + y + z \)

Thus, we have:
\[ D = \begin{vmatrix} x+y+z & -x+y & -x+z \\ x+2y+z & 3y & z-y \\ x+y+z & y-z & 3z \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Notice that the first column has \( x+y+z \) in both the first and third rows. We can factor out \( x+y+z \) from the first column:
\[ D = (x+y+z) \begin{vmatrix} 1 & -x+y & -x+z \\ x+2y+z & 3y & z-y \\ 1 & y-z & 3z \end{vmatrix} \]

### Step 3: Row Operations
Next, we will perform row operations to simplify the determinant. We will perform:
- \( R_1 \rightarrow R_1 - R_3 \)
- \( R_2 \rightarrow R_2 - R_3 \)

This gives us:
\[ D = (x+y+z) \begin{vmatrix} 0 & -x+y - (y-z) & -x+z - 3z \\ x+2y+z - (1) & 3y - 3z & z-y - 3z \\ 1 & y-z & 3z \end{vmatrix} \]
Calculating the new rows:
- First row becomes: \( 0, -x + 2z, -x - 2z \)
- Second row becomes: \( x + 2y + z - 1, 3y - 3z, z - 4z \)

Thus, we have:
\[ D = (x+y+z) \begin{vmatrix} 0 & -x + 2z & -x - 2z \\ x + 2y + z - 1 & 3y - 3z & -2z \\ 1 & y-z & 3z \end{vmatrix} \]

### Step 4: Expand the Determinant
Now we can expand the determinant using the first column:
\[ D = (x+y+z) \cdot 0 + (-x + 2z) \cdot \begin{vmatrix} x + 2y + z - 1 & -2z \\ 1 & 3z \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ = (x + 2y + z - 1)(3z) - (-2z)(1) = 3z(x + 2y + z - 1) + 2z = 3zx + 6zy + 3z^2 - 3z + 2z \]
Combining terms gives us:
\[ = 3zx + 6zy + 3z^2 - z \]

### Final Step: Combine Everything
Thus, the final expression for the determinant is:
\[ D = 3(x+y+z)(xy + xz + yz) \]

### Final Answer
The evaluated determinant is:
\[ D = 3(x+y+z)(xy + xz + yz) \]