JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 25)
Let a line passing through the point $$(-1,2,3)$$ intersect the lines $$L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$$ at $$M(\alpha, \beta, \gamma)$$ and $$L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$$ at $$N(a, b, c)$$. Then, the value of $$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$$ equals __________.
Answer
196
Explanation
$$\begin{aligned} & \mathrm{M}(3 \lambda+1,2 \lambda+2,-2 \lambda-1) \quad \therefore \alpha+\beta+\gamma=3 \lambda+2 \\ & \mathrm{~N}(-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\mu+1 \end{aligned}$$
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$$\begin{aligned} & \frac{3 \lambda+2}{-3 \mu-1}=\frac{2 \lambda}{-2 \mu}=\frac{-2 \lambda-4}{4 \mu-2} \\ & 3 \lambda \mu+2 \mu=3 \lambda \mu+\lambda \\ & 2 \mu=\lambda \\ & 2 \lambda \mu-\lambda=\lambda \mu+2 \mu \\ & \lambda \mu=\lambda+2 \mu \\ &\Rightarrow \lambda \mu=2 \lambda \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad & \mu=2 \quad(\lambda \neq 0) \\ \therefore \quad & \lambda=4 \\ & \alpha+\beta+\gamma=14 \\ & a+b+c=-1 \\ & \frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}=196 \end{aligned}$$
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