Question

Let $P$ be the 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. $205\sqrt{5}$
• C. $17/\sqrt{5}$
• D. $11/\sqrt{5}$

Answer 1

Answer: (D)

$11/\sqrt{5}$

$\lambda (x+y+z-6)+2x+3y+z+5=0$
$(\lambda +2)x+(\lambda +3)y+(\lambda +1)z+5-6\lambda =0\lambda +1=0\Rightarrow \lambda =-1$
$P:x+2y+11=0$
perpendicular distance $=\frac{11}{\sqrt{5}}$