A) \[\frac{3}{2}\]
B) \[\frac{9}{2}\]
C) \[-\frac{2}{9}\]
D) \[-\frac{3}{2}\]
Correct Answer: B
Solution :
We have the lines \[\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\] ?(i) and \[\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\] ?(ii) Let a point \[\left( 2r+1,\,\,3r-1,\,\,4r+1 \right)\]be on the line Eq. (i). If this is an intersection point of both the lines, then it will lie on Eq. (ii), also \[\therefore \] \[\frac{2r+1-3}{1}=\frac{3r-1-k}{2}=\frac{4r+1}{1}\] ?(iii) Taking first and third part of Eq. (iii), we get \[2r-2=4r+1\] \[\Rightarrow \] \[r=-\frac{3}{2}\] Taking second and third part of Eq. (iii), we get \[3r-1-k=8r+2\] \[\Rightarrow \] \[3r-1-k-8r-2=0\] \[\Rightarrow \] \[k=-5r-3\] \[\Rightarrow \] \[k=-5\left( -\frac{3}{2} \right)-3\] \[\left( \because \,r=-\frac{3}{2} \right)\] \[\Rightarrow \] \[k=\frac{15}{2}-3\] \[\Rightarrow \] \[k=\frac{9}{2}\]
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