Let the line $\mathrm{L}$ intersect the lines $x-2=-y=z-1,2(x+1)=2(y-1)=z+1$ 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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Aicharya · Accepted Answer

$\mathrm{L}_1: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\lambda$ $\mathrm{L}_2: \frac{\mathrm{x}+1}{\frac{1}{2}}=\frac{\mathrm{y}-1}{\frac{1}{2}}=\frac{\mathrm{z}+1}{1}=\mu$ dr of line MN will be $ < 3+\lambda-\frac{\mu}{2},-1-\lambda-\frac{\mu}{2}, 2+\lambda-\mu>\&$ it will be proportional to $ < 3,1,2>$

$\Rightarrow \lambda=-\frac{4}{3} \& \mu=-\frac{2}{3}$ $\therefore$ Coordinate of $\mathrm{M}$ will be $ < \left(\frac{2}{3}, \frac{4}{3},-\frac{1}{3}\right)$ and equation of required line will be. $\frac{\mathrm{x}-\frac{2}{3}}{3}=\frac{\mathrm{y}-\frac{4}{3}}{1}=\frac{\mathrm{z}+\frac{1}{3}}{2}=\mathrm{k}$ So any point on this line will be $\begin{aligned} & \left(\frac{2}{3}+3 \mathrm{k}, \frac{4}{3}+\mathrm{k},-\frac{1}{3}+2 \mathrm{k}\right) \\ & \because \frac{2}{3}+3 \mathrm{k}=-\frac{1}{3} \Rightarrow \mathrm{k}=-\frac{1}{3} \end{aligned}$ $\therefore$ Point lie on the line for $\mathrm{k}=-\frac{1}{3} \text { is }\left(-\frac{1}{3}, 1,-1\right)$

Let the line $\mathrm{L}$ intersect the lines $x-2=-y=z-1,2(x+1)=2(y-1)=z+1$ and be parallel to the line…

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