### Three Dimensional Geometry Question 33

#### Question 33 - 01 February - Shift 2

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 :

(1) $\frac{250}{83}$

(2) $\frac{15}{53}$

(3) $\frac{25}{83}$

(4) $\frac{250}{82}$

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Family of Planes, A Plane and A Point

$P \equiv P_1+\lambda P_2=0$

$(2+\lambda) x+(3+2 \lambda) y+(3 \lambda-1) z-2-6 \lambda=0$

Plane $P$ is perpendicular to $P_3 $

$\therefore \overrightarrow{{}n} \cdot \overrightarrow{{}n}_3=0$

$2(\lambda+2)+(2 \lambda+3)-(3 \lambda-1)=0$

$\lambda=-8$

$P \equiv-6 x-13 y-25 z+46=0$

$6 x+13 y+25 z-46=0$

Distance from $(-7,1,1)$

$d=|\frac{-42+13+25-46}{\sqrt{36+169+625}}|=\frac{50}{\sqrt{830}}$

$d^{2}=\frac{50 \times 50}{830}=\frac{250}{83}$