`|{:(1+a,c,1+bc),(1+a,b,1+bc),(1+a,e,1+bc):}|` is equal to

Options

A. 1
B. 0
C. 3
D. `a+b+c`

Answer

To solve the determinant \( D = \begin{vmatrix} 1+a & c & 1+bc \\ 1+a & b & 1+bc \\ 1+a & e & 1+bc \end{vmatrix} \), we can follow these steps:

### Step 1: Write down the determinant
We start with the given determinant:
\[ D = \begin{vmatrix} 1+a & c & 1+bc \\ 1+a & b & 1+bc \\ 1+a & e & 1+bc \end{vmatrix} \]

### Step 2: Simplify the determinant
Notice that the first column has the same entry \( 1+a \) in all three rows. This suggests that we can perform row operations to simplify the determinant. We can subtract the first row from the second and third rows.

Perform the following operations:
- \( R_2 \rightarrow R_2 - R_1 \)
- \( R_3 \rightarrow R_3 - R_1 \)

This gives us:
\[ D = \begin{vmatrix} 1+a & c & 1+bc \\ 0 & b-c & 0 \\ 0 & e-c & 0 \end{vmatrix} \]

### Step 3: Factor out common terms
Now we can see that the second and third rows have zeros in the first column. We can factor out the common terms from the second and third rows:
\[ D = (1+a) \begin{vmatrix} 1 & c & 1+bc \\ 0 & b-c & 0 \\ 0 & e-c & 0 \end{vmatrix} \]

### Step 4: Evaluate the determinant
The determinant can be evaluated using the property that if any two rows (or columns) of a determinant are identical, the determinant is zero. Here, since the second and third rows are multiples of each other (both have zeros in the first column), we conclude:
\[ D = 0 \]

### Final Result
Thus, the value of the determinant is:
\[ \boxed{0} \]