JEE Main 2024 — 27 January, Shift 2 — Question 24

Given lines: $L_1: 2x - y + 3 = 0, \quad$ $L_2: 6x + 3y + 1 = 0, \quad$ $L_3: \alpha x + 2y - 2 = 0$

Step 1: Condition for lines not forming a triangle:

• Two lines parallel or all three concurrent.

Step 2: Check concurrency using determinant:

$$\begin{vmatrix} 2 & -1 & 3 \\ 6 & 3 & 1 \\ \alpha & 2 & -2 \end{vmatrix} = 0$$

Compute determinant:

$$2\begin{vmatrix}3 & 1\\ 2 & -2\end{vmatrix} -(-1)\begin{vmatrix}6 &1\\ \alpha &-2\end{vmatrix} + 3\begin{vmatrix}6 &3\\ \alpha &2\end{vmatrix} = 0$$ $$2(-8) -(-12-\alpha) + 3(12-3\alpha) = 0$$ $$-16 -12 -\alpha +36 -9\alpha =0 \Rightarrow -10\alpha +8=0 \Rightarrow \alpha = \frac{4}{5}$$

Step 3: Check for parallel cases:

• $L_1 \parallel L_3 \Rightarrow 2/\alpha = -1/2 \Rightarrow \alpha = -4$
• $L_2 \parallel L_3 \Rightarrow 6/\alpha = 3/2 \Rightarrow \alpha = 4$

Step 4: Sum of squares of all $\alpha$:

$$p = \left(\frac{4}{5}\right)^2 + 4^2 + (-4)^2 = \frac{16}{25} + 16 +16 = \frac{816}{25} = 32.64$$

Greatest integer less than or equal to $p$:

$$\lfloor p \rfloor = \boxed{32}$$