### Three Dimensional Geometry Question 30

#### Question 30 - 31 January - Shift 2

Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$. Then the distance of $P$ from the point $(3,-2,2)$ is

(1) 6

(2) 4

(3) 5

(4) 7

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: A Plane and A Point (7.1)

Equation of Plane :

$2(x-1)-3(y+1)-6(z+5)=0$

Or $2 x-3 y-6 z=35$

$\Rightarrow$ Required distan ce $=$

$\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}$

$\sqrt{4+9+36}$

$=5$