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The plane 4x+7y+4z+81=0 is rotated throu...

The plane `4x+7y+4z+81=0` is rotated through a right angle about its line of intersection with the plane `5x+3y+10 z=25.` The equation of the plane in its new position is `x-4y+6z=k` where `k` is

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The correct Answer is:
`7`
`" "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)+ lamda(5x+3y+10z-25)=0`
or `" "(4+5lamda)x+ (7+3lamda)y+ (4+10lamda)z+ 81- 25lamda=0" "` (iii)
Plane (iii) `bot` to (i), so
`" "4(4+5lamda)+7(7+3lamda)+ 4(4+10lamda)=0`
`therefore" "lamda=-1`
From (iii), equation of plane is
`" "-x+4y-6z+106=0" "`(iv)
Distance of (iv) from (0, 0, 0)`=(106)/(sqrt(1+16+ 36))=(106)/(sqrt(53))`
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