For all real values of a and b lines `(2a + b) x + (a + 3b) y + (b - 3a) =0 and mx + 2y + 6=0` are concurent, then m is equal to

Options

A. `-2`
B. `-3`
C. `-4`
D. `-5`

Answer

To determine the value of \( m \) such that the lines \( (2a + b)x + (a + 3b)y + (b - 3a) = 0 \) and \( mx + 2y + 6 = 0 \) are concurrent for all real values of \( a \) and \( b \), we can follow these steps:

### Step 1: Write the equations of the lines in standard form
The first line can be expressed as:
\[ L_1: (2a + b)x + (a + 3b)y + (b - 3a) = 0 \]
The second line is:
\[ L_2: mx + 2y + 6 = 0 \]

### Step 2: Identify the coefficients
From the first line \( L_1 \), we have:
- Coefficient of \( x \): \( 2a + b \)
- Coefficient of \( y \): \( a + 3b \)
- Constant term: \( b - 3a \)

From the second line \( L_2 \), we have:
- Coefficient of \( x \): \( m \)
- Coefficient of \( y \): \( 2 \)
- Constant term: \( 6 \)

### Step 3: Use the condition for concurrency
For the lines to be concurrent, the determinant formed by the coefficients must be zero. The determinant is given by:
\[ \begin{vmatrix} 2a + b & a + 3b & b - 3a \\ m & 2 & 6 \\ \end{vmatrix} = 0 \]

This can be expanded as:
\[ (2a + b)(2 \cdot 6) - (a + 3b)(m) + (b - 3a)(m) = 0 \]

### Step 4: Calculate the determinant
Expanding the determinant:
\[ 12(2a + b) - m(a + 3b) + m(b - 3a) = 0 \]
This simplifies to:
\[ 12(2a + b) - m(a + 3b - b + 3a) = 0 \]
\[ 12(2a + b) - m(4a + 2b) = 0 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 12(2a + b) = m(4a + 2b) \]
Dividing both sides by \( 2 \):
\[ 12a + 6b = 2ma + mb \]

### Step 6: Grouping terms
Rearranging terms:
\[ (12 - 2m)a + (6 - m)b = 0 \]

### Step 7: Setting coefficients to zero
For this equation to hold for all \( a \) and \( b \), both coefficients must be zero:
1. \( 12 - 2m = 0 \)
2. \( 6 - m = 0 \)

### Step 8: Solving the equations
From \( 12 - 2m = 0 \):
\[ 2m = 12 \implies m = 6 \]
From \( 6 - m = 0 \):
\[ m = 6 \]

### Conclusion
Thus, the value of \( m \) is:
\[ \boxed{6} \]