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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 $k$ is equal to
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The correct answer is:
$\frac{9}{2}$
$\frac{9}{2}$
Any point on $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=t$ is $(2 t+1,3 t-1,4 t+1)$
And any point on $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mathrm{s}$ is $(s+3,2 s+k, s)$
Given lines are intersecting
$\Rightarrow \mathrm{t}=-\frac{3}{2} \text { and } \mathrm{s}=-5 \quad \therefore \mathrm{k}=\frac{9}{2}$
And any point on $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mathrm{s}$ is $(s+3,2 s+k, s)$
Given lines are intersecting
$\Rightarrow \mathrm{t}=-\frac{3}{2} \text { and } \mathrm{s}=-5 \quad \therefore \mathrm{k}=\frac{9}{2}$
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