Three Dimensional Geometry Question 19
Question 19 - 2024 (30 Jan Shift 2)
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 $N(a, b, \mathrm{c})$. Then the value of $\frac{(\alpha+\beta+\gamma)^{2}}{(a+b+c)^{2}}$ equals
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Answer (196)
Solution
$\mathrm{M}(3 \lambda+1,2 \lambda+2,-2 \lambda-1) \quad \therefore \alpha+\beta+\gamma=3 \lambda+2$
$\mathrm{N}(-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\mu+1$
$$ \begin{aligned} & \frac{3 \lambda+2}{-3 \mu-1}=\frac{2 \lambda}{-2 \mu}=\frac{-2 \lambda-4}{4 \mu-2} \ & 3 \lambda \mu+2 \mu=3 \lambda \mu+\lambda \ & 2 \mu=\lambda \ & 2 \lambda \mu-\lambda=\lambda \mu+2 \mu \ & \lambda \mu=\lambda+2 \mu \ \Rightarrow & \lambda \mu=2 \lambda \ \Rightarrow & \mu=2 \quad(\lambda \neq 0) \ \therefore & \lambda=4 \ & \alpha+\beta+\gamma=14 \ & a+b+c=-1 \ & \frac{(\alpha+\beta+\gamma)^{2}}{(a+b+c)^{2}}=196 \end{aligned} $$
