For what value of `k` are the three st.lines :
`2x+y-3=0` , `5x+ky-3=0` and `3x-y-2=0` are concurrent.

Answer

To find the value of \( k \) for which the three straight lines \( 2x + y - 3 = 0 \), \( 5x + ky - 3 = 0 \), and \( 3x - y - 2 = 0 \) are concurrent, we will use the condition that the determinant of the coefficients of these lines must be equal to zero.

### Step-by-Step Solution:

1. **Identify the coefficients of the lines**:
- For the first line \( 2x + y - 3 = 0 \), the coefficients are \( a_1 = 2 \), \( b_1 = 1 \), \( c_1 = -3 \).
- For the second line \( 5x + ky - 3 = 0 \), the coefficients are \( a_2 = 5 \), \( b_2 = k \), \( c_2 = -3 \).
- For the third line \( 3x - y - 2 = 0 \), the coefficients are \( a_3 = 3 \), \( b_3 = -1 \), \( c_3 = -2 \).

2. **Set up the determinant**:
The determinant \( D \) for the three lines is given by:
\[ D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = \begin{vmatrix} 2 & 1 & -3 \\ 5 & k & -3 \\ 3 & -1 & -2 \end{vmatrix} \]

3. **Calculate the determinant**:
Expanding the determinant using the first row:
\[ D = 2 \begin{vmatrix} k & -3 \\ -1 & -2 \end{vmatrix} - 1 \begin{vmatrix} 5 & -3 \\ 3 & -2 \end{vmatrix} - 3 \begin{vmatrix} 5 & k \\ 3 & -1 \end{vmatrix} \]

Now, calculate each of the 2x2 determinants:
- \( \begin{vmatrix} k & -3 \\ -1 & -2 \end{vmatrix} = k \cdot (-2) - (-3) \cdot (-1) = -2k - 3 \)
- \( \begin{vmatrix} 5 & -3 \\ 3 & -2 \end{vmatrix} = 5 \cdot (-2) - (-3) \cdot 3 = -10 + 9 = -1 \)
- \( \begin{vmatrix} 5 & k \\ 3 & -1 \end{vmatrix} = 5 \cdot (-1) - k \cdot 3 = -5 - 3k \)

Substitute these back into the determinant:
\[ D = 2(-2k - 3) - 1(-1) - 3(-5 - 3k) \]
Simplifying this gives:
\[ D = -4k - 6 + 1 + 15 + 9k \]
Combine like terms:
\[ D = (9k - 4k) + (-6 + 1 + 15) = 5k + 10 \]

4. **Set the determinant to zero**:
For the lines to be concurrent, we set \( D = 0 \):
\[ 5k + 10 = 0 \]

5. **Solve for \( k \)**:
\[ 5k = -10 \\ k = -2 \]

### Final Answer:
The value of \( k \) for which the three lines are concurrent is \( k = -2 \).