The value of k such that the lines `2x-3y+k=0,3x-4y-13=0` and `8x-11y-33=0` are concurrent is

Options

A. 20
B. `-7`
C. `7`
D. `-20`

Explanation

To find the value of \( k \) such that the lines \( 2x - 3y + k = 0 \), \( 3x - 4y - 13 = 0 \), and \( 8x - 11y - 33 = 0 \) are concurrent, we can use the determinant method. The lines are concurrent if the determinant of the coefficients of \( x \), \( y \), and the constant terms is zero.

### Step-by-step Solution:

1. **Write down the equations of the lines:**
\[ L_1: 2x - 3y + k = 0 \quad (1) \]
\[ L_2: 3x - 4y - 13 = 0 \quad (2) \]
\[ L_3: 8x - 11y - 33 = 0 \quad (3) \]

2. **Form the determinant using the coefficients:**
The determinant is formed as follows:
\[ D = \begin{vmatrix} 2 & -3 & k \\ 3 & -4 & -13 \\ 8 & -11 & -33 \end{vmatrix} \]

3. **Calculate the determinant:**
We can expand the determinant:
\[ D = 2 \begin{vmatrix} -4 & -13 \\ -11 & -33 \end{vmatrix} - (-3) \begin{vmatrix} 3 & -13 \\ 8 & -33 \end{vmatrix} + k \begin{vmatrix} 3 & -4 \\ 8 & -11 \end{vmatrix} \]

Now, we calculate each of the 2x2 determinants:
- For the first determinant:
\[ \begin{vmatrix} -4 & -13 \\ -11 & -33 \end{vmatrix} = (-4)(-33) - (-13)(-11) = 132 - 143 = -11 \]
- For the second determinant:
\[ \begin{vmatrix} 3 & -13 \\ 8 & -33 \end{vmatrix} = (3)(-33) - (-13)(8) = -99 + 104 = 5 \]
- For the third determinant:
\[ \begin{vmatrix} 3 & -4 \\ 8 & -11 \end{vmatrix} = (3)(-11) - (-4)(8) = -33 + 32 = -1 \]

4. **Substituting back into the determinant:**
\[ D = 2(-11) + 3(5) + k(-1) = -22 + 15 - k \]
\[ D = -7 - k \]

5. **Set the determinant equal to zero for concurrency:**
\[ -7 - k = 0 \]

6. **Solve for \( k \):**
\[ k = -7 \]

### Conclusion:
The value of \( k \) such that the lines are concurrent is \( k = -7 \).