Let a line passing through the point (–1, 2, 3) intersect the lines at and at N(a, b, c). Then the value of equals [2024]

Question

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 $M (α, β, γ)$ and $L_{2} : \frac{x + 2}{- 3} = \frac{y - 2}{- 2} = \frac{z - 1}{4}$ at N(a, b, c). Then the value of $\frac{{(α + β + γ)}^{2}}{{(a + b + c)}^{2}}$ equals __________. [2024]

Smart Achievers · Accepted Answer

(196)

We have, $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{- 2} = λ \in R$

$∴ M \equiv (3 λ + 1, 2 λ + 2, - 2 λ - 1)$

$L_{2} : \frac{x + 2}{- 3} = \frac{y - 2}{- 2} = \frac{z - 1}{4} = μ \in R$

$∴ N \equiv (- 3 μ — 2, - 2 μ + 2, 4 μ + 1)$

$α + β + γ = 3 λ + 2 \Rightarrow a + b + c = - μ + 1$

Direction ratio's of line AM = $< 3 λ + 2, 2 λ, - 2 λ - 4 >$

Direction ratio's of line AN = $< - 3 μ — 1, - 2 μ, 4 μ - 2 >$

$\Rightarrow \frac{3 λ + 2}{- 3 μ - 1} = \frac{2 λ}{- 2 μ} = \frac{- 2 λ - 4}{4 μ - 2}$

$\Rightarrow \frac{3 λ + 2}{- 3 μ - 1} = \frac{λ}{- μ} and \frac{λ}{- μ} = \frac{- 2 λ - 4}{4 μ - 2}$

$\Rightarrow - 3 λ μ - 2 μ = - 3 μ λ - λ and 4 μ λ - 2 λ = 2 μ λ + 4 μ$

$\Rightarrow 2 μ = λ and 2 μ λ = 2 λ + 4 μ$

$\Rightarrow λ^{2} = 2 λ + 2 λ \Rightarrow λ^{2} = 4 λ$

$\Rightarrow λ (λ - 4) = 0 \Rightarrow λ = 0, λ = 4$

$∴ \frac{{(α + β + γ)}^{2}}{{(a + b + c)}^{2}} = \frac{{(3 \times 4 + 2)}^{2}}{{(- 2 + 1)}^{2}} = \frac{{(14)}^{2}}{1} = 196$ .

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Q. Let a line passing through the point (–1, 2, 3) intersect the lines L1:x–13=y–22=z+1–2 at M(α,β,γ) and L2:x+2–3=y–2–2=z–14 at N(a, b, c). Then the value of (α+β+γ)2(a+b+c)2 equals __________. [2024]

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