Question

Let the plane $P$ pass through the intersection of the planes $2 x+3 y-z=2$ and $x+2 y+3 z=6$, and be perpendicular to the plane $2 x+y-z+1=0$. 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-6 x-13 y-25 z+46=0 $
$ 6 x+13 y +25 z -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}$