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Question

The plane denoted by 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 is denoted by P, and the distance of this plane from the origin is k, then the value of [k2] is (where [.] represents greatest integer less than or equal to k).

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Solution

4x+7y+4z+81=0 (i)5x+3y+10z=25 (ii)
Equation of plane passing through their line of intersection is
4x+7y+4z+81+λ(5x+3y+10z25)=0
or (4+5λ)x+(7+3λ)y+(4+10λ)z+8125λ=0 (iii)
plane (iii) perpendicular to (i), so
4(4+5λ)+7(7+3λ)+4(4+10λ)=0λ=1
From (iii), equation of the plane is
x+4y6z+106=0 (iv)
Distance of (iv) from (0,0,0)=1061+16+36=10653

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