Question

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

Answer 1

$4x+7y+4z+81=0....\left(i\right)$
$5x+3y+10z=25....\left(ii\right)$
Equation of plane passing through their line of intersection is
$(4x+7y+4z+81)+\lambda (5x+3y+10z-25)=0$
or $(4+5\lambda )x+(7+3\lambda )y+(4+10\lambda )z+81-25\lambda =0........\left(iii\right)$

Plane $\left(iii\right)\perp$ to $\left(i\right)$, so
$4(4+5\lambda )+7(7+3\lambda )+4(4+10\lambda )=0\Rightarrow \lambda =-1$

From$\left(iii\right),$ equation of plane is $-x+4y-6z+106=0....\left(iv\right)$

Distance of $\left(iv\right)\text{from}(0,0,0)$
$d=\frac{106}{\sqrt{1+16+36}}=\frac{106}{\sqrt{53}}$
$\Rightarrow \left[\frac{d}{2}\right]=7$