Let the line L intersect the lines x-2=-y=z-1 | vedclass

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(B) Let the line $L$ intersect the line $L_1: \frac{x-2}{1} = \frac{y}{-1} = \frac{z-1}{1} = \lambda$ at point $M(2+\lambda, -\lambda, 1+\lambda)$ and

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$\left(-\frac{1}{3}, 1, -1\right)$

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(B) Let the line $L$ intersect the line $L_1: \frac{x-2}{1} = \frac{y}{-1} = \frac{z-1}{1} = \lambda$ at point $M(2+\lambda, -\lambda, 1+\lambda)$ and the line $L_2: \frac{x+1}{1/2} = \frac{y-1}{1/2} = \frac{z+1}{1} = \mu$ at point $N(-1+\mu/2, 1+\mu/2, -1+\mu)$.Since $L$ is parallel to the line with direction ratios $\langle 3, 1, 2 \rangle$,the vector $\vec{MN}$ must be parallel to $\langle 3, 1, 2 \rangle$.$\vec{MN} = \langle (-1+\mu/2) - (2+\lambda), (1+\mu/2) - (-\lambda), (-1+\mu) - (1+\lambda) \rangle = \langle \mu/2 - \lambda - 3, \mu/2 + \lambda + 1, \mu - \lambda - 2 \rangle$.Since $\vec{MN} \parallel \langle 3, 1, 2 \rangle$,we have $\frac{\mu/2 - \lambda - 3}{3} = \frac{\mu/2 + \lambda + 1}{1} = \frac{\mu - \lambda - 2}{2}$.From $\frac{\mu/2 + \lambda + 1}{1} = \frac{\mu - \lambda - 2}{2}$,we get $\mu + 2\lambda + 2 = \mu - \lambda - 2 \Rightarrow 3\lambda = -4 \Rightarrow \lambda = -4/3$.From $\frac{\mu/2 - \lambda - 3}{3} = \frac{\mu/2 + \lambda + 1}{1}$,we get $\mu/2 - \lambda - 3 = 3\mu/2 + 3\lambda + 3 \Rightarrow -\mu = 4\lambda + 6$. Substituting $\lambda = -4/3$,we get $-\mu = 4(-4/3) + 6 = -16/3 + 18/3 = 2/3 \Rightarrow \mu = -2/3$.Point $M = (2 - 4/3, 4/3, 1 - 4/3) = (2/3, 4/3, -1/3)$.The equation of line $L$ is $\frac{x-2/3}{3} = \frac{y-4/3}{1} = \frac{z+1/3}{2} = k$.Any point on $L$ is $(2/3 + 3k, 4/3 + k, -1/3 + 2k)$.Setting $2/3 + 3k = -1/3 \Rightarrow 3k = -1 \Rightarrow k = -1/3$.For $k = -1/3$,the point is $(2/3 - 1, 4/3 - 1/3, -1/3 - 2/3) = (-1/3, 1, -1)$.

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

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