The value of the determinant `|(10!,11!,12!),(11!,12!,13!),(12!,13!,14!)|`, is

Options

A. `2(10! 11!)`
B. `2(10! 13!)`
C. `2(10! 11! 12!)`
D. `2(11 ! 12! 13!)`

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} 10! & 11! & 12! \\ 11! & 12! & 13! \\ 12! & 13! & 14! \end{vmatrix} \]

we can simplify the factorials in the determinant.

### Step 1: Factor out common terms from each row

Notice that we can express the factorials in terms of \(10!\):

- \(11! = 11 \times 10!\)
- \(12! = 12 \times 11!\)
- \(13! = 13 \times 12!\)
- \(14! = 14 \times 13!\)

Thus, we can rewrite the determinant as:

\[ D = \begin{vmatrix} 10! & 11 \times 10! & 12 \times 11 \times 10! \\ 11 \times 10! & 12 \times 11 \times 10! & 13 \times 12 \times 11 \times 10! \\ 12 \times 11 \times 10! & 13 \times 12 \times 11 \times 10! & 14 \times 13 \times 12 \times 11 \times 10! \end{vmatrix} \]

### Step 2: Factor out \(10!\) from the first row, \(11!\) from the second row, and \(12!\) from the third row

Now we can factor out \(10!\), \(11!\), and \(12!\) from each row respectively:

\[ D = 10! \cdot 11! \cdot 12! \cdot \begin{vmatrix} 1 & 11 & 12 \times 11 \\ 11 & 12 \times 11 & 13 \times 12 \\ 12 \times 11 & 13 \times 12 & 14 \times 13 \end{vmatrix} \]

### Step 3: Simplify the determinant

Now we need to compute the determinant:

\[ D' = \begin{vmatrix} 1 & 11 & 12 \times 11 \\ 11 & 12 \times 11 & 13 \times 12 \\ 12 \times 11 & 13 \times 12 & 14 \times 13 \end{vmatrix} \]

### Step 4: Perform row operations

We can perform row operations to simplify the determinant. Let's subtract the first row from the second and third rows:

- \(R_2 \rightarrow R_2 - 11R_1\)
- \(R_3 \rightarrow R_3 - 12 \times 11R_1\)

This gives us:

\[ D' = \begin{vmatrix} 1 & 11 & 12 \times 11 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{vmatrix} \]

### Step 5: Calculate the determinant

Now we can calculate the determinant:

\[ D' = 1 \cdot (1 \cdot 2 - 1 \cdot 0) = 2 \]

### Step 6: Combine the results

Now substituting back into the expression for \(D\):

\[ D = 10! \cdot 11! \cdot 12! \cdot 2 \]

Thus, the final value of the determinant is:

\[ D = 2 \cdot 10! \cdot 11! \cdot 12! \]

### Final Answer

The value of the determinant is:

\[ \boxed{2 \cdot 10! \cdot 11! \cdot 12!} \]