Let the line L pass through (1,1,1) and inter | vedclass

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(D) Let the line $L$ pass through $C(1,1,1)$.Let the line $L$ intersect the line $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$ at point $

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$(7,15,13)$

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(D) Let the line $L$ pass through $C(1,1,1)$.Let the line $L$ intersect the line $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$ at point $A(2\lambda+1, 3\lambda-1, 4\lambda+1)$.Let the line $L$ intersect the line $L_2: \frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1} = \mu$ at point $B(\mu+3, 2\mu+4, \mu)$.Since $A, B, C$ are collinear,the direction ratios of $AC$ and $BC$ must be proportional.Direction ratios of $AC$ are $(2\lambda+1-1, 3\lambda-1-1, 4\lambda+1-1) = (2\lambda, 3\lambda-2, 4\lambda)$.Direction ratios of $BC$ are $(\mu+3-1, 2\mu+4-1, \mu-1) = (\mu+2, 2\mu+3, \mu-1)$.Since $A, B, C$ are collinear,$\frac{2\lambda}{\mu+2} = \frac{3\lambda-2}{2\mu+3} = \frac{4\lambda}{\mu-1} = k$.From $\frac{2\lambda}{\mu+2} = \frac{4\lambda}{\mu-1}$,we get $\frac{1}{\mu+2} = \frac{2}{\mu-1} \Rightarrow \mu-1 = 2\mu+4 \Rightarrow \mu = -5$.Substituting $\mu = -5$ into the coordinates of $B$,we get $B(-5+3, 2(-5)+4, -5) = (-2, -6, -5)$.The direction ratios of line $L$ (passing through $C(1,1,1)$ and $B(-2, -6, -5)$) are $(1-(-2), 1-(-6), 1-(-5)) = (3, 7, 6)$.The equation of line $L$ is $\frac{x-1}{3} = \frac{y-1}{7} = \frac{z-1}{6}$.Checking the options:For $(7, 15, 13)$: $\frac{7-1}{3} = 2, \frac{15-1}{7} = 2, \frac{13-1}{6} = 2$. Since all ratios are equal,$(7, 15, 13)$ lies on line $L$.

Let the line $L$ pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$. Then,which of the following points lies on the line $L$?

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