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If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then $k=$
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The correct answer is:
$\frac{9}{2}$
Given two lines intersect. Thus shortest distance between them is zero.
Let $\left(x_{1}, y_{1}, z_{1}\right) \equiv(1,-1,1)$ and $\left(x_{2}, y_{2}, z_{2}\right) \equiv(3, k, 0)$
Let $\left(x_{1}, y_{1}, z_{1}\right) \equiv(1,-1,1)$ and $\left(x_{2}, y_{2}, z_{2}\right) \equiv(3, k, 0)$
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