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Question

The plane P1:4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane P2:5x+3y+10z=25. If the plane in its new position be denoted by P and the distance of plane P from the origin is d units, then the value of [d/2], where [.] represents the greatest integer function, is

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Solution

P1:4x+7y+4z+81=0
P2:5x+3y+10z=25

Equation of plane passing through line of intersection of P1 and P2 is
P:(4x+7y+4z+81)+λ(5x+3y+10z25)=0
(4+5λ)x+(7+3λ)y+(4+10λ)z+8125λ=0
This plane is perpendicular to P1.
So 4(4+5λ)+7(7+3λ)+4(4+10λ)=0
λ=1
Hence, equation of the plane P is x+4y6z+106=0

Distance of plane P from (0,0,0) is
d=1061+16+36=10653
Thus, [d/2]=7

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