JEE Mains · Maths · STD 12 - 11. three dimension geometry
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 \(\mathrm{M}(\alpha, \beta, \gamma)\) and \(\mathrm{L}_2: \frac{\mathrm{x}+2}{-3}=\frac{\mathrm{y}-2}{-2}=\frac{\mathrm{z}-1}{4}\) at \(\mathrm{N}(\mathrm{a}, \mathrm{b}\), c). Then the value of \(\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}\) equals
- A \(100\)
- B \(196\)
- C \(150\)
- D \(190\)
Answer & Solution
Correct Answer
(B) \(196\)
Step-by-step Solution
Detailed explanation
\(\mathrm{M}(3 \lambda+1,2 \lambda+2,-2 \lambda-1) \quad \therefore \alpha+\beta+\gamma=3 \lambda+2 \) \(\mathrm{~N}(-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\mu+1\)…
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