The determinant
`Delta=|(al+a'l',am+a'm',an+a'n'),(bl+b'l',bm+b'm',bn+b'n'),(cl+c'l',cm+c'm',cn+c'n')|` is equal to

Options

A. `(abc+a'b'c)(lmn+l'm'n')`
B. `abclmn+a'b'c'l'm'n'`
C. `(a^(2)+b^(2)+c^(2))(l^(2)+m^(2)+n^(2))+(a^('2)+b^('2)+c^('2))(l^('2)+m^('2)+n^('2))`
D. 0

Answer

To solve the determinant

\[ \Delta = \begin{vmatrix} al + a'l' & am + a'm' & an + a'n' \\ bl + b'l' & bm + b'm' & bn + b'n' \\ cl + c'l' & cm + c'm' & cn + c'n' \end{vmatrix} \]

we can rewrite it in a more manageable form.

### Step 1: Rewrite the determinant
Notice that each entry in the determinant can be expressed as a sum of products. We can represent the determinant as a product of two determinants.

Let:
\[ A = \begin{vmatrix} a & a' \\ b & b' \\ c & c' \end{vmatrix} \quad \text{and} \quad B = \begin{vmatrix} l & l' \\ m & m' \\ n & n' \end{vmatrix} \]

Then, we can express \(\Delta\) as:
\[ \Delta = A \cdot B \]

### Step 2: Analyze the structure
The determinant \(A\) has the third column as zeros (since there are no terms involving \(l\), \(m\), or \(n\) in the last column). Similarly, the determinant \(B\) will also have a row of zeros.

### Step 3: Identify the zero rows/columns
Since one column of \(A\) (the third column) is entirely zero, the value of \(A\) is zero:
\[ A = 0 \]

Similarly, since one row of \(B\) (the third row) is entirely zero, the value of \(B\) is also zero:
\[ B = 0 \]

### Step 4: Compute the determinant
Thus, we have:
\[ \Delta = A \cdot B = 0 \cdot 0 = 0 \]

### Conclusion
The value of the determinant \(\Delta\) is:
\[ \Delta = 0 \]