If the line drawn from the point `(-2,-1,-3)` meets a plane at right angle at the point `(1,-3,3),` find the equation of the plane.

Options

A. ` 3x+2y+6z-18=0`
B. ` 3x+2y+6z-24=0`
C. ` 3x-2y+6z-24=0`
D. ` 3x-2y+6z-27=0`

Answer

D

Explanation

Since the line drawn from the point `(-2,-1,-3)` meets a plane at right angle at the point `(1,-3,3)`. So the plane passes through the point `(1,-3,3)` and normal to plane is `(-3hati+2hatj-5hatk)` ltrbgt `implies veca=hati-3hatj+3hatk`
and `vecN=-3hati+2hatj-6hatk`
So the equation of required plane is `(vecr-veca).vecN=0`
`implies[(xhati+yhatj+zhatk)-(hati-3hatj+3hatk)].(-3hatk+2hatj-6hatk)=0`
`implies [(x-1)hati+(y+3)hatj+(z-3)hatk].(-3hati+2hatj-6hatk)=0`
`implies -3x+3+2y+6-6z+18=0`
`implies -3x+2y-6z=-27`
`:. 3x-2y+6z-27=0`