If `|(a,b,c),(l,m,n),(p,q,r)|=2`, then what is the value of the determinant `|(6a,3b,15c),(2l,m,5n),(2p,q,5r)|` = ?

Options

A. 10
B. 20
C. 40
D. 60

Answer

To solve the determinant \( |(6a, 3b, 15c), (2l, m, 5n), (2p, q, 5r)| \) given that \( |(a, b, c), (l, m, n), (p, q, r)| = 2 \), we can use properties of determinants.

### Step-by-Step Solution:

1. **Identify the original determinant**:
We know that:
\[ |(a, b, c), (l, m, n), (p, q, r)| = 2 \]

2. **Factor out constants from the rows**:
The determinant can be expressed in terms of the original determinant by factoring out the constants from each row:
\[ |(6a, 3b, 15c), (2l, m, 5n), (2p, q, 5r)| = 6 \cdot 3 \cdot 15 \cdot |(a, b, c), (l, m, n), (p, q, r)| \]
Here, we factor out:
- 6 from the first row,
- 3 from the second row,
- 15 from the first row.

3. **Calculate the product of the constants**:
Calculate the product of the constants factored out:
\[ 6 \cdot 3 \cdot 15 = 270 \]

4. **Multiply by the original determinant**:
Now, multiply this result by the value of the original determinant:
\[ |(6a, 3b, 15c), (2l, m, 5n), (2p, q, 5r)| = 270 \cdot 2 = 540 \]

5. **Adjust for the constants factored out from the second and third rows**:
We also need to factor out the constants from the second and third rows:
- From the second row, we factor out \(2\) (from \(2l, m, 5n\)).
- From the third row, we factor out \(5\) (from \(2p, q, 5r\)).

Thus, we have:
\[ |(6a, 3b, 15c), (2l, m, 5n), (2p, q, 5r)| = 270 \cdot 2 \cdot 5 \cdot 2 = 540 \]

6. **Final calculation**:
Therefore, the final value of the determinant is:
\[ |(6a, 3b, 15c), (2l, m, 5n), (2p, q, 5r)| = 60 \]

### Conclusion:
The value of the determinant is \( 60 \).