A) 11
B) \[6\sqrt{11}\]
C) \[\frac{11}{\sqrt{6}}\]
D) \[11\sqrt{6}\]
Correct Answer: C
Solution :
Let\[{{L}_{1}}:\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}\] and\[{{L}_{2}}:\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}\] \[\therefore \]Equation of plane containing \[{{L}_{1}}\]and \[{{L}_{2}}\]is \[\left| \begin{matrix} x-1 & y-4 & z+4 \\ 3 & 5 & 7 \\ 1 & 4 & 7 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[(x-1)(35-28)-(y-4)(21-7)+(z+4)(12-5)=0\] \[\Rightarrow \]\[7x-14y+7z+77=0\Rightarrow x-2y+z+11=0\] \[\therefore \]Perpendicular distance from the origin to the plane\[\frac{|11|}{\sqrt{1+4+1}}=\frac{11}{\sqrt{6}}\]units
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