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If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to _________
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Verified Answer
The correct answer is:
43
$\begin{aligned} & \overline{\mathrm{a}}_1=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overline{\mathrm{a}}_2=-2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{p}}-=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{q}}-=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & (\lambda+2) \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}=\overline{\mathrm{a}}_1-\overline{\mathrm{a}}_2 \\ & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}-=-6 \hat{\mathrm{i}}-15 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\end{aligned}$
$\begin{aligned}
& \frac{44}{\sqrt{30}}=\frac{|-6 \lambda-12-105-9|}{\sqrt{(-6)^2+(-15)^2+3^2}} \\
& \frac{44}{\sqrt{30}}=\frac{|6 \lambda+126|}{3 \sqrt{30}} \\
& 132=|6 \lambda+126| \\
& \lambda=1, \lambda=-43 \\
& |\lambda|=43
\end{aligned}$
$\begin{aligned}
& \frac{44}{\sqrt{30}}=\frac{|-6 \lambda-12-105-9|}{\sqrt{(-6)^2+(-15)^2+3^2}} \\
& \frac{44}{\sqrt{30}}=\frac{|6 \lambda+126|}{3 \sqrt{30}} \\
& 132=|6 \lambda+126| \\
& \lambda=1, \lambda=-43 \\
& |\lambda|=43
\end{aligned}$
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