If `6a^2-3b^2-c^2+7ab-ac+4bc=0` then the family of lines `ax+by+c=0,|a|+|b| != 0` can be concurrent at concurrent (A) (-2,3) (B) (3,-1) (C) (2,3) (D) (-3,1)

Options

A. `(-2,-3)`
B. `(3,-1)`
C. `(2,3)`
D. `(-3,1)`

Explanation

To solve the given equation \( 6a^2 - 3b^2 - c^2 + 7ab - ac + 4bc = 0 \) and determine the concurrent points of the family of lines \( ax + by + c = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ 6a^2 + 7ab - 3b^2 - ac + 4bc - c^2 = 0 \]
This is a quadratic equation in terms of \( a \).

### Step 2: Identifying Coefficients
In the quadratic equation \( Ax^2 + Bx + C = 0 \), we identify:
- \( A = 6 \)
- \( B = 7b - c \)
- \( C = -3b^2 + 4bc - c^2 \)

### Step 3: Finding the Discriminant
For the quadratic equation to have real and concurrent solutions, the discriminant must be equal to zero:
\[ D = B^2 - 4AC = 0 \]
Substituting the values:
\[ (7b - c)^2 - 4(6)(-3b^2 + 4bc - c^2) = 0 \]

### Step 4: Expanding the Discriminant
Expanding the discriminant:
\[ (7b - c)^2 + 72b^2 - 96bc + 24c^2 = 0 \]
This simplifies to:
\[ 49b^2 - 14bc + c^2 + 72b^2 - 96bc + 24c^2 = 0 \]
Combining like terms:
\[ 121b^2 - 110bc + 25c^2 = 0 \]

### Step 5: Solving the Quadratic in \( b \) and \( c \)
Now, we treat this as a quadratic in \( b \):
\[ 121b^2 - 110bc + 25c^2 = 0 \]
Using the quadratic formula \( b = \frac{-B \pm \sqrt{D}}{2A} \):
Where \( A = 121 \), \( B = -110c \), and \( C = 25c^2 \).

### Step 6: Finding the Roots
The discriminant \( D \) for this quadratic is:
\[ D = (-110c)^2 - 4 \cdot 121 \cdot 25c^2 = 12100c^2 - 12100c^2 = 0 \]
This indicates that \( b \) has a double root.

### Step 7: Finding the Concurrent Points
Setting \( b \) in terms of \( c \):
\[ b = \frac{110c}{2 \cdot 121} = \frac{55c}{121} \]

### Step 8: Substituting Back to Find Points
Substituting back into the line equation \( ax + by + c = 0 \):
1. For \( b = 3 \) and \( c = -1 \):
- \( 3A - B + C = 0 \) gives the point \( (3, -1) \).

2. For \( b = 2 \) and \( c = 3 \):
- \( 2A + 3B - C = 0 \) gives the point \( (2, 3) \).

3. For \( b = -2 \) and \( c = 3 \):
- \( -2A + 3B - C = 0 \) gives the point \( (-2, 3) \).

4. For \( b = -3 \) and \( c = 1 \):
- \( -3A + B - C = 0 \) gives the point \( (-3, 1) \).

### Conclusion
The concurrent points from the options provided are:
- (A) (-2, 3)
- (B) (3, -1)
- (C) (2, 3)
- (D) (-3, 1)

The correct answer is **(B) (3, -1)**.