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Question

The distance of the point (1,3,7) from the plane passing through the point (1,1,1), having normal perpendicular to both the lines x11=y+22=z43 and x22=y+11=z+71, is:

A
2074
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B
1083
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C
583
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D
1074
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Solution

The correct option is B 1083
As the plane is parallel to both the lines. So the cross product of direction ratios of the lines will give the normal vector of the plane
(1,2,3)×(2,1,1)=(5,7,3)
The equation of plane with normal and a point is
r.n=a.n(ra).n=0.
So the equation of plane passing through the point (1,1,1) and direction ratios (5,7,3) is

5(x1)+7(y+1)+3(z+1)=0
Hence, 5x+7y+3z+5=0 is the equation of the plane. So, the distance of the point P(1,3,7) from the plane is
d=5+2121+525+49+9=1083units

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