Question

Let the plane $P$ pass through the intersection of the planes $2x+3y-z=2$ and $x+2y+3z=6$, and be perpendicular to the plane $2x+y-z+1=0$. If $d$ is the distance of $P$ from the point $(-7,1,1)$, then $d^2$ is equal to :

• A. $\frac{25}{83}$
• B. $\frac{250}{83}$
• C. $\frac{15}{53}$
• D. $\frac{250}{82}$

Answer 1

Answer: (B)

$P\equiv P_1+\lambda P_2=0$
$(2+\lambda )x+(3+2\lambda )y+(3\lambda -1)z-2-6\lambda =0$
$\text{ Plane }P\text{ is perpendicular to }{P}_3$
$\therefore \vec{n}\cdot {\vec{n}}_3=0$
$2(\lambda +2)+(2\lambda +3)-(3\lambda -1)=0$
$\lambda =-8$
$P\equiv -6x-13y-25z+46=0$
$6x+13y+25z-46=0$
$\text{ Dist from }(-7,1,1)$
$d=\left|\frac{-42+13+25-46}{\sqrt{36+169+625}}\right|=\frac{50}{\sqrt{830}}$
$d^2=\frac{50\times 50}{830}=\frac{250}{83}$