Question

The plane $P_1:4x+7y+4z+81=0$ is rotated through a right angle about its line of intersection with the plane $P_2:5x+3y+10z=25.$ 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 $\left[d/2\right],$ where $[.]$ represents the greatest integer function, is

Answer 1

$P_1:4x+7y+4z+81=0$
$P_2:5x+3y+10z=25$

Equation of plane passing through line of intersection of $P_1$ and $P_2$ is
$P:(4x+7y+4z+81)+\lambda (5x+3y+10z-25)=0$
$\Rightarrow (4+5\lambda )x+(7+3\lambda )y+(4+10\lambda )z+81-25\lambda =0$
This plane is perpendicular to $P_1.$
So $4(4+5\lambda )+7(7+3\lambda )+4(4+10\lambda )=0$
$\Rightarrow \lambda =-1$
Hence, equation of the plane $P$ is $-x+4y-6z+106=0$

Distance of plane $P$ from $(0,0,0)$ is
$d=\frac{106}{\sqrt{1+16+36}}=\frac{106}{\sqrt{53}}$
Thus, $\left[d/2\right]=7$