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The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is
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$3 \sqrt{30}$ units
$\begin{aligned} & \text { Shortest distance }=\left|\frac{\left|\begin{array}{ccc}3-(-3) & 8-(-7) & 3-6 \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|}{\sqrt{(-4-2)^2+(12+3)^2+(6-3)^2}}\right| \\ & =\frac{|6 \times(-6)-15 \times(15)-3 \times(3)|}{\sqrt{36+225+9}} \\ & =\frac{270}{\sqrt{270}}=\sqrt{270}=3 \sqrt{30}\end{aligned}$
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