If the lines `x - 2y - 6 = 0 , 3x + y - 4=0 ` and `lambda x +4y + lambda^(2) = 0 ` are concurrent , then value of `lambda` is

Answer

To find the value of \(\lambda\) for which the lines \(x - 2y - 6 = 0\), \(3x + y - 4 = 0\), and \(\lambda x + 4y + \lambda^2 = 0\) are concurrent, we can use the condition that the determinant of the coefficients of these lines must be zero.

### Step-by-step Solution:

1. **Identify the coefficients of the lines**:
- For the line \(x - 2y - 6 = 0\):
- \(a_1 = 1\), \(b_1 = -2\), \(c_1 = -6\)
- For the line \(3x + y - 4 = 0\):
- \(a_2 = 3\), \(b_2 = 1\), \(c_2 = -4\)
- For the line \(\lambda x + 4y + \lambda^2 = 0\):
- \(a_3 = \lambda\), \(b_3 = 4\), \(c_3 = \lambda^2\)

2. **Set up the determinant**:
The lines are concurrent if the determinant of the coefficients is zero:
\[ \begin{vmatrix} 1 & -2 & -6 \\ 3 & 1 & -4 \\ \lambda & 4 & \lambda^2 \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
Expanding the determinant using the first row:
\[ = 1 \cdot \begin{vmatrix} 1 & -4 \\ 4 & \lambda^2 \end{vmatrix} - (-2) \cdot \begin{vmatrix} 3 & -4 \\ \lambda & \lambda^2 \end{vmatrix} - 6 \cdot \begin{vmatrix} 3 & 1 \\ \lambda & 4 \end{vmatrix} \]

Now calculating each of these 2x2 determinants:
- First determinant:
\[ = 1 \cdot (1 \cdot \lambda^2 - (-4) \cdot 4) = \lambda^2 + 16 \]
- Second determinant:
\[ = 2 \cdot (3 \cdot \lambda^2 - (-4) \cdot \lambda) = 2(3\lambda^2 + 4\lambda) = 6\lambda^2 + 8\lambda \]
- Third determinant:
\[ = -6 \cdot (3 \cdot 4 - 1 \cdot \lambda) = -6(12 - \lambda) = -72 + 6\lambda \]

4. **Combine the results**:
Now we combine all these results:
\[ \lambda^2 + 16 + 6\lambda^2 + 8\lambda - 72 + 6\lambda = 0 \]
Simplifying this gives:
\[ 7\lambda^2 + 14\lambda - 56 = 0 \]

5. **Divide by 7**:
\[ \lambda^2 + 2\lambda - 8 = 0 \]

6. **Factor the quadratic**:
Factoring gives:
\[ (\lambda + 4)(\lambda - 2) = 0 \]

7. **Find the values of \(\lambda\)**:
Setting each factor to zero gives:
\[ \lambda + 4 = 0 \quad \Rightarrow \quad \lambda = -4 \]
\[ \lambda - 2 = 0 \quad \Rightarrow \quad \lambda = 2 \]

### Final Answer:
The values of \(\lambda\) for which the lines are concurrent are \(\lambda = 2\) and \(\lambda = -4\).

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