JEE Advanced 2024 — Paper 1 — Question 12

We are given $\vec{OP} = \frac{\alpha-1}{\alpha} \hat{i} + \hat{j} + \hat{k}$, $\vec{OQ} = \hat{i} + \frac{\beta-1}{\beta} \hat{j} + \hat{k}$, and $\vec{OR} = \hat{i} + \hat{j} + \frac{1}{2} \hat{k}$. The condition $(\vec{OP} \times \vec{OQ}) \cdot \vec{OR} = 0$ is the scalar triple product, which equals the determinant $\begin{vmatrix} \frac{\alpha-1}{\alpha} & 1 & 1 \\ 1 & \frac{\beta-1}{\beta} & 1 \\ 1 & 1 & \frac{1}{2} \end{vmatrix} = 0$. Expanding along the first row: $\frac{\alpha-1}{\alpha} \left( \frac{\beta-1}{\beta} \cdot \frac{1}{2} - 1 \cdot 1 \right) - 1 \left( 1 \cdot \frac{1}{2} - 1 \cdot 1 \right) + 1 \left( 1 \cdot 1 - \frac{\beta-1}{\beta} \cdot 1 \right) = 0$. Simplify: $\frac{\alpha-1}{\alpha} \left( \frac{\beta-1}{2\beta} - 1 \right) - \left( \frac{1}{2} - 1 \right) + \left( 1 - \frac{\beta-1}{\beta} \right) = 0$ → $\frac{\alpha-1}{\alpha} \cdot \frac{\beta-1-2\beta}{2\beta} - \left(-\frac{1}{2}\right) + \frac{\beta - (\beta-1)}{\beta} = 0$ → $\frac{\alpha-1}{\alpha} \cdot \frac{-\beta-1}{2\beta} + \frac{1}{2} + \frac{1}{\beta} = 0$. Multiply through by $2\alpha\beta$: $-(\alpha-1)(\beta+1) + \alpha\beta + 2\alpha = 0$ → $-(\alpha\beta + \alpha - \beta - 1) + \alpha\beta + 2\alpha = 0$ → $-\alpha\beta - \alpha + \beta + 1 + \alpha\beta + 2\alpha = 0$ → $\beta + \alpha + 1 = 0$.

Hence $\alpha + \beta = -1$. The point $(\alpha, \beta, 2)$ lies on the plane $3x+3y-z+l=0$, so $3\alpha + 3\beta - 2 + l = 0$ → $3(\alpha+\beta) - 2 + l = 0$.

Substitute $\alpha+\beta = -1$: $3(-1) - 2 + l = 0$ → $-3 - 2 + l = 0$ → $l = 5$.