Question

The plane denoted by $P_1:4x+7y+4z+81=0$ 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 is denoted by $P$, and the distance of this plane from the origin is $k$, then the value of $\left[\frac{k}{2}\right]$ is (where [.] represents greatest integer less than or equal to $k$).

Answer 1

$4x+7y+4z+81=0\cdots \left(i\right)5x+3y+10z=25\cdots \left(ii\right)$
Equation of plane passing through their line of intersection is
$4x+7y+4z+81+\lambda (5x+3y+10z-25)=0$
or $(4+5\lambda )x+(7+3\lambda )y+(4+10\lambda )z+81-25\lambda =0\cdots \left(iii\right)$
plane $\left(iii\right)$ perpendicular to $\left(i\right)$, so
$4(4+5\lambda )+7(7+3\lambda )+4(4+10\lambda )=0\therefore \lambda =-1$
From $\left(iii\right)$, equation of the plane is
$-x+4y-6z+106=0\cdots \left(iv\right)$
Distance of $\left(iv\right)$ from $(0,0,0)=\frac{106}{\sqrt{1+16+36}}=\frac{106}{\sqrt{53}}$