If `Delta =|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|` then the value of `|(2a_1+3b_1+4c_1,b_1,c_1),(2a2+3b_2+4c_2,b_2,c_2),(2a_3+3b_3+4c_3,b_3,c_3)|` is equal to

Options

A. `Delta^2`
B. `4Delta`
C. `Delta`
D. `2Delta`

Answer

To solve the given problem, we start with the determinant:

\[ \Delta = |(a_1, b_1, c_1), (a_2, b_2, c_2), (a_3, b_3, c_3)| \]

We need to find the value of:

\[ |(2a_1 + 3b_1 + 4c_1, b_1, c_1), (2a_2 + 3b_2 + 4c_2, b_2, c_2), (2a_3 + 3b_3 + 4c_3, b_3, c_3)| \]

### Step 1: Write the determinant

We can write the determinant explicitly as follows:

\[ D = |(2a_1 + 3b_1 + 4c_1, b_1, c_1), (2a_2 + 3b_2 + 4c_2, b_2, c_2), (2a_3 + 3b_3 + 4c_3, b_3, c_3)| \]

### Step 2: Split the first column

We can split the first column of the determinant into three separate terms:

\[ D = |(2a_1, b_1, c_1), (2a_2, b_2, c_2), (2a_3, b_3, c_3)| + |(3b_1, b_1, c_1), (3b_2, b_2, c_2), (3b_3, b_3, c_3)| + |(4c_1, b_1, c_1), (4c_2, b_2, c_2), (4c_3, b_3, c_3)| \]

### Step 3: Factor out constants

Now, we can factor out the constants from each of the three determinants:

\[ D = 2 |(a_1, b_1, c_1), (a_2, b_2, c_2), (a_3, b_3, c_3)| + 3 |(b_1, b_1, c_1), (b_2, b_2, c_2), (b_3, b_3, c_3)| + 4 |(c_1, b_1, c_1), (c_2, b_2, c_2), (c_3, b_3, c_3)| \]

### Step 4: Analyze the determinants

1. The first determinant simplifies to \(2 \Delta\).
2. The second determinant has two identical columns (the second column is the same as the first column), which makes it equal to 0.
3. The third determinant also has two identical columns (the first and third columns are the same), which also makes it equal to 0.

Thus, we have:

\[ D = 2 \Delta + 0 + 0 = 2 \Delta \]

### Conclusion

Therefore, the value of the determinant is:

\[ \boxed{2\Delta} \]