Lines $2x+y-1=0$, $ax+3y-3=0$ and $3x+2y-2=0$ will be concurrent:

Question

Accepted Answer

To determine the concurrency of the lines given by the equations $2x + y - 1 = 0$, $ax + 3y - 3 = 0$, and $3x + 2y - 2 = 0$, we need to find a common point of intersection for all three lines. Lines are concurrent if they intersect at a single point.

The condition for three lines $a_1x + b_1y + c_1 = 0$, $a_2x + b_2y + c_2 = 0$, and $a_3x + b_3y + c_3 = 0$ to be concurrent is that the determinant of their coefficients must be zero:

$$
\begin{vmatrix}
a_1 & b_1 & c_1 \
a_2 & b_2 & c_2 \
a_3 & b_3 & c_3 \
\end{vmatrix}
= 0
$$

For the given lines, the coefficients are:
- Line 1: $a_1 = 2$, $b_1 = 1$, $c_1 = -1$
- Line 2: $a_2 = a$, $b_2 = 3$, $c_2 = -3$
- Line 3: $a_3 = 3$, $b_3 = 2$, $c_3 = -2$

The determinant is:

$$
\begin{vmatrix}
2 & 1 & -1 \
a & 3 & -3 \
3 & 2 & -2 \
\end{vmatrix}
= 2(3 \cdot -2 - (-3) \cdot 2) - 1(a \cdot -2 - (-3) \cdot 3) - 1(a \cdot 2 - 3 \cdot 3)
$$

Calculating each part:

1. $2(3 \cdot -2 - (-3) \cdot 2) = 2(-6 + 6) = 2 \cdot 0 = 0$
2. $-1(a \cdot -2 - (-3) \cdot 3) = -1(-2a + 9) = 2a - 9$
3. $-1(a \cdot 2 - 3 \cdot 3) = -1(2a - 9) = -2a + 9$

Adding these results gives:

$$0 + (2a - 9) + (-2a + 9) = 0$$

This simplifies to:

$$0 + 0 = 0$$

The determinant is zero for all real values of $a$, which means the lines are concurrent for any real value of $a$.

**Explanation of Incorrect Options:**

- **Option 2 (for $a=4$ only):** This option suggests concurrency only at $a=4$. However, as shown, the determinant is zero for all $a$, not just $a=4$.

- **Option 3 (for $-1 \le a \le 3$ only):** This option restricts concurrency to a specific range of $a$. The determinant calculation shows concurrency is not limited to this range but holds for all real $a$.

- **Option 4 (for $a>0$ only):** This option suggests concurrency only for positive $a$. Again, the determinant is zero for all real $a$, including negative and zero values.

Thus, Option 1 is correct because the lines are concurrent for all real values of $a$.