Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to___________

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{P}(-3 \lambda-2,4 \lambda+2,2 \lambda+5) \\ & \mathrm{Q}(-\mu-2,2 \mu-6,1) \\ & \mathrm{DRS} \text { of } \mathrm{PQ}=(3 \lambda-\mu, 2 \mu-4 \lambda-8,-2 \lambda-4) \\ & \mathrm{DRS} \text { of } \mathrm{PQ}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & 2 & 0 \\ -3 & 4 & 2 \end{array}\right| \\ & =(4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \end{aligned}$ OR $\begin{aligned} & (2,1,1) \\ & \frac{3 \lambda-\mu}{2}=\frac{2 \mu-4 \lambda-8}{1}=\frac{-2 \lambda-4}{1} \\ & \Rightarrow \mu=\lambda+2 \& 7 \lambda=\mu-8 \\ & \lambda=-1 \quad \mu=1 \\ & Q:(-3,-4,1) \\ & L_{P Q}=\frac{x+3}{2}=\frac{y+4}{1}=\frac{z-1}{1} \\ & (-1, \alpha, \beta) \Rightarrow 1=\frac{\alpha+4}{1}=\frac{\beta-1}{1} \\ & \Rightarrow \alpha=-3, \beta=2 \\ & (\alpha-\beta)^2=25 \end{aligned}$

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines…

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