The value of `|{:(1+a,b,c),(a,1+b,c),(a,b,1+c):}|` is

Options

A. abc
B. `a+b+c`
C. `1+a+b+c`
D. `3+abc`

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 1 + a & b & c \\ a & 1 + b & c \\ a & b & 1 + c \end{vmatrix} \]

we will perform row operations to simplify the determinant.

### Step 1: Apply Row Operations
We will perform the following row operations:
1. \( R_2 \to R_2 - R_1 \)
2. \( R_3 \to R_3 - R_1 \)

This will help us simplify the determinant.

### Step 2: Calculate the New Rows
After performing the operations:

- For \( R_2 \):
\[ R_2 = (a - (1 + a), (1 + b) - b, c - c) = (0, 1, 0) \]

- For \( R_3 \):
\[ R_3 = (a - (1 + a), b - b, (1 + c) - c) = (0, 0, 1) \]

So now, the determinant becomes:

\[ D = \begin{vmatrix} 1 + a & b & c \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Now, we can calculate the determinant using the property of determinants:

\[ D = (1 + a) \cdot \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} - b \cdot \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} + c \cdot \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} \]

The determinant of the \(2 \times 2\) identity matrix is \(1\), and the other two determinants are \(0\):

\[ D = (1 + a) \cdot 1 - b \cdot 0 + c \cdot 0 = 1 + a \]

### Step 4: Final Result
Thus, we have:

\[ D = 1 + a + b + c \]

### Conclusion
The value of the determinant is:

\[ \boxed{1 + a + b + c} \]