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Let $L_{1}$ and $L_{2}$ be the following straight lines $L_{1} : \frac{x - 1}{1} = \frac{y}{- 1} = \frac{z - 1}{3}$ and $L_{2} : \frac{x - 1}{- 3} = \frac{y}{- 1} = \frac{z - 1}{1}$ suppose the straight line $L : \frac{x - α}{l} = \frac{y - 1}{m} = \frac{z - γ}{- 2}$ Lies in the plane containing $L_{1}$ and $L_{2}$ and passes through the point of intersection of $L_{1}$ and $L_{2}$ If the line L bisect the acute angle between the lines $L_{1}$ and $L_{2}$ then which of the following statements is/are TRUE

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$α - γ = 1$

$l + m = 2$

$l + m = 0$

$α - γ = 3$

answer is A, B.

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Detailed Solution

$∴ Let C = (λ_{1} + 1.2 λ_{1} + 2.3 λ_{1} + 3)$

$and D = (λ_{2} + 1.2 λ_{2} + 2.3 λ_{2} + 3)$

$As, C D = \sqrt{14}$

$\Rightarrow {(λ_{1} - λ_{2})}^{2} + 4 {(λ_{1} - λ_{2})}^{2} + 9 {(λ_{1} - λ_{2})}^{2} = 14$

$\Rightarrow λ_{2} = 1 + λ_{1}$

$let G = (α, β, γ)$

$∴ 4 α = 5 + 2 λ_{1}$

$4 β = 10 + 4 λ_{1}$

$4 γ = 9 + 6 λ_{1}$

$The locus of G = (α, β, γ) is \frac{4 x - 5}{2}$

$= \frac{4 y - 10}{4} = \frac{4 z - 9}{6} = λ (say)$

$∴$ Any point on it is

$(4 x = 2 λ + 5, 4 y = 4 λ + 10.4 z = 6 λ + 9)$

But, we are given $(\frac{3}{2}, y_{1}, z_{1})$

$∴ (\frac{3}{2}) = 2 λ + 5 \Rightarrow λ = \frac{1}{2} ∴ (\frac{3}{2} \cdot y_{1} = 3, z_{1} = 3)$

$\Rightarrow Locus is \frac{(x - \frac{3}{2})}{1} = \frac{(y - 3)}{2} = \frac{z - 3}{2}$

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