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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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Verified Answer
The correct answer is:
25

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