Let $\overrightarrow{O P}=\frac{\alpha-1}{\alpha} \hat{i}+\hat{j}+\hat{k}, \overrightarrow{O…

Let $\overrightarrow{O P}=\frac{\alpha-1}{\alpha} \hat{i}+\hat{j}+\hat{k}, \overrightarrow{O Q}=\hat{i}+\frac{\beta-1}{\beta} \hat{j}+\hat{k}$ and $\overrightarrow{O R}=\hat{i}+\hat{j}+\frac{1}{2} \hat{k}$ be three vectors, where $\alpha, \beta \in \mathbb{R}-\{0\}$ and $O$ denotes the origin. If $(\overrightarrow{O P} \times \overrightarrow{O Q}) \cdot \overrightarrow{O R}=0$ and the point $(\alpha, \beta, 2)$ lies on the plane $3 x+3 y-z+l=0$, then the value of $l$ is ____

Solution

$(\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OQ}}) \cdot \overrightarrow{\mathrm{OR}}=0$ $\left|\begin{array}{ccc}\frac{\alpha-1}{\alpha} & 1 & 1 \\ 1 & \frac{\beta-1}{\beta} & 1 \\ 1 & 1 & \frac{1}{2}\end{array}\right|=0$ $\alpha+\beta+1=0...(1)$ Also $\quad(\alpha, \beta, 2)$ lies on $3 \mathrm{x}+3 \mathrm{y}-\mathrm{z}+l=0$ $\Rightarrow \quad 3 \alpha+3 \beta-2+l=0 \quad \Rightarrow \quad l=2-3(\alpha+\beta)$ ...(2) use (1) in (2) $\Rightarrow l=5$

Asked in: JEE Advanced 2024 (Paper 1)

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