A) 10
B) 9
C) 7
D) 6
Correct Answer: D
Solution :
Given equation of the line is \[\frac{x-3}{0}=\frac{y-7}{1}=\frac{z-1}{1}=(k)\] (say) Any point on the line is \[(3,\,7+k,\,1+k)\] The line joining \[A(5,\,\,1,\,\,3)\] and \[B(3,\,7+k,\,\,1+k)\] is perpendicular to the given line if \[0\,(-2)\,+\,(6+k)\cdot \,(1)\,+(k-2)\,\cdot \,(1)=0\] \[\Rightarrow \] \[6+k+k-2=0\] \[\Rightarrow \] \[2k+4=0\] \[\Rightarrow \] \[k=-2\] \[\therefore \] Any point B(3, 5, - 1) lie on the line. \[\therefore \] Required distance \[AB=\sqrt{{{2}^{2}}+{{4}^{2}}+{{4}^{2}}}\] \[=\sqrt{36}=6\]
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