The value(s) of t for which the lines `2x+3y=5, t^(2)x+ty-6=0 and 3x-2y-1=0` are concurrent, can be :

Options

A. `t=2`
B. `t=-3`
C. `t=-2`
D. `t=3`

Answer

To find the value(s) of \( t \) for which the lines \( 2x + 3y = 5 \), \( t^2 x + ty - 6 = 0 \), and \( 3x - 2y - 1 = 0 \) are concurrent, we will use the condition for concurrency of three lines. The lines are concurrent if the determinant of their coefficients equals zero.

### Step-by-Step Solution:

1. **Identify the coefficients of the lines:**
- For the first line \( 2x + 3y - 5 = 0 \), the coefficients are:
\[ a_1 = 2, \quad b_1 = 3, \quad c_1 = -5 \]
- For the second line \( t^2 x + ty - 6 = 0 \), the coefficients are:
\[ a_2 = t^2, \quad b_2 = t, \quad c_2 = -6 \]
- For the third line \( 3x - 2y - 1 = 0 \), the coefficients are:
\[ a_3 = 3, \quad b_3 = -2, \quad c_3 = -1 \]

2. **Set up the determinant:**
The condition for concurrency is given by the determinant:
\[ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \]
Plugging in the coefficients, we have:
\[ \begin{vmatrix} 2 & 3 & -5 \\ t^2 & t & -6 \\ 3 & -2 & -1 \end{vmatrix} = 0 \]

3. **Calculate the determinant:**
Expanding the determinant:
\[ = 2 \begin{vmatrix} t & -6 \\ -2 & -1 \end{vmatrix} - 3 \begin{vmatrix} t^2 & -6 \\ 3 & -1 \end{vmatrix} - 5 \begin{vmatrix} t^2 & t \\ 3 & -2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
- \( \begin{vmatrix} t & -6 \\ -2 & -1 \end{vmatrix} = t(-1) - (-6)(-2) = -t - 12 \)
- \( \begin{vmatrix} t^2 & -6 \\ 3 & -1 \end{vmatrix} = t^2(-1) - (-6)(3) = -t^2 + 18 \)
- \( \begin{vmatrix} t^2 & t \\ 3 & -2 \end{vmatrix} = t^2(-2) - t(3) = -2t^2 - 3t \)

Substituting back into the determinant:
\[ 2(-t - 12) - 3(-t^2 + 18) - 5(-2t^2 - 3t) = 0 \]
Simplifying this:
\[ -2t - 24 + 3t^2 - 54 + 10t^2 + 15t = 0 \]
Combine like terms:
\[ 13t^2 + 13t - 78 = 0 \]

4. **Factor the quadratic equation:**
Dividing the entire equation by 13:
\[ t^2 + t - 6 = 0 \]
Factoring:
\[ (t + 3)(t - 2) = 0 \]

5. **Find the values of \( t \):**
Setting each factor to zero gives:
\[ t + 3 = 0 \quad \Rightarrow \quad t = -3 \]
\[ t - 2 = 0 \quad \Rightarrow \quad t = 2 \]

### Final Answer:
The values of \( t \) for which the lines are concurrent are \( t = -3 \) and \( t = 2 \).