JEE Main 2025 Jan 24 Shift 1, Mathematics Q19: Points and Straight Lines

Step 1: Find $\lambda$ from the image condition. The midpoint of $(1,2)$ and $(57/13,-40/13)$ lies on the mirror line:

Midpoint $=\left(\frac{57/13+1}{2},\frac{-40/13+2}{2}\right)=\left(\frac{35}{13},-\frac{7}{13}\right)$

Substitute into $2x-3y+\lambda =0$:

$2\cdot \frac{35}{13}-3\cdot \left(-\frac{7}{13}\right)+\lambda =0\Rightarrow \frac{70+21}{13}+\lambda =0\Rightarrow \lambda =-7$.

Step 2: Concurrency $\Rightarrow$ determinant of coefficients is zero.

$\left|\begin{array}{c}3-4-\alpha \\ 8-11-33\\ 2-3\lambda \end{array}\right|=0$

Expand along row 1:

$3(-11\lambda -99)+4(8\lambda +66)-\alpha (-24+22)=0$

$-33\lambda -297+32\lambda +264+2\alpha =0\Rightarrow -\lambda -33+2\alpha =0$

With $\lambda =-7$: $7-33+2\alpha =0\Rightarrow \alpha =13$.

$|\alpha \lambda |=|13\cdot (-7)|=91$.

$\therefore$ Option (2).