Find the value of `a` for which the lines `2x+y-1=0`, `a x+3y-3=0`, `3x+2y-2=0` are concurrent.

Options

A. -3
B. -1
C. 1
D. infinite value

Explanation

To find the value of \( a \) for which the lines \( 2x + y - 1 = 0 \), \( ax + 3y - 3 = 0 \), and \( 3x + 2y - 2 = 0 \) are concurrent, we can use the determinant method. The lines are concurrent if the determinant of the coefficients of the equations is zero.

### Step 1: Write the equations in standard form
The equations are:
1. \( 2x + y - 1 = 0 \)
2. \( ax + 3y - 3 = 0 \)
3. \( 3x + 2y - 2 = 0 \)

### Step 2: Set up the determinant
The determinant of the coefficients of \( x \), \( y \), and the constant terms is given by:

\[ \begin{vmatrix} 2 & 1 & -1 \\ a & 3 & -3 \\ 3 & 2 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
We can calculate the determinant using the formula for a 3x3 matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, this becomes:

\[ D = 2(3 \cdot (-2) - (-3) \cdot 2) - 1(a \cdot (-2) - (-3) \cdot 3) + (-1)(a \cdot 2 - 3 \cdot 3) \]

Calculating each term:
1. \( 2(3 \cdot (-2) - (-3) \cdot 2) = 2(-6 + 6) = 2(0) = 0 \)
2. \( -1(a \cdot (-2) - (-3) \cdot 3) = -1(-2a + 9) = 2a - 9 \)
3. \( -1(a \cdot 2 - 3 \cdot 3) = -1(2a - 9) = -2a + 9 \)

Combining these, we get:

\[ D = 0 + (2a - 9) + (-2a + 9) = 0 \]

### Step 4: Set the determinant to zero
Since the lines are concurrent, we set the determinant equal to zero:

\[ 0 = 0 \]

This means that the determinant does not depend on \( a \) and is always zero.

### Conclusion
Thus, the lines are concurrent for any value of \( a \). Therefore, \( a \) can be any real number.