Question

Let $P$ be a plane, which contains the line of intersection of the planes, $x+y+z-6=0$ and $2x+3y+z+5=0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0,0,256)$ from $P$ is equal to :

• A. $63\sqrt{5}$
• B. $\frac{11}{\sqrt{5}}$
• C. $\frac{205}{\sqrt{5}}$
• D. $\frac{17}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{11}{\sqrt{5}}$

Equation of plane $P$ is :
$(x+y+z-6)+\lambda (2x+3y+z+5)=0$
$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1+\lambda )z+(-6+5\lambda )=0$
Since, Plane $P$ is perpendicular to $xy$-plane.
$\therefore 1+\lambda =0\Rightarrow \lambda =-1$
Hence, equation of plane $P:x+2y+11=0$
Perpendicular distance of plane from $(0,0,256)=\frac{11}{\sqrt{5}}$