JEE Main 2024 — 1 February, Shift 2 — Question 15

$\overrightarrow{\mathrm{PN}} \cdot \overrightarrow{\mathrm{b}}=0$ ?

$3(3 \lambda-2)+2(2 \lambda-5)+(\lambda-7)=0$

$14 \lambda=23 \Rightarrow \lambda=\frac{23}{14}$

$\mathrm{N}\left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right)$

$\therefore \frac{\alpha+3}{2}=\frac{83}{14} \Rightarrow \alpha=\frac{62}{7}$

$\frac{\beta+4}{2}=\frac{32}{14} \Rightarrow \beta=\frac{4}{7}$

$\frac{\gamma+9}{2}=\frac{51}{14} \Rightarrow \gamma=\frac{-12}{7}$

Ans. $14(\alpha+\beta+r)=108$