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)

$\frac{250}{83}$

Answer 2

P≡P1​+λP2​=0
(2+λ)x+(3+2λ)y+(3λ−1)z−2−6λ=0
Plane P is perpendicular to P3​
∴n⋅n3​=0
2(λ+2)+(2λ+3)−(3λ−1)=0
λ=−8
P≡−6x−13y−25z+46=0
6x+13y+25z−46=0
Dist from (−7,1,1)
d=∣∣​36+169+625​−42+13+25−46​∣∣​=830​50​
d2=83050×50​=83250​