Answer :

Given: Plane is perpendicular to and is at a distance of 5 units from origin.

To find: Equation of plane

Formula Used: Equation of a plane is lx + my + nz = p where p is the distance from the origin and l, m and n are the direction cosines of the normal to the plane

Explanation:

Direction ratio of normal to the plane is 2:-3:1

Therefore, direction cosines of the normal to the plane is

l , m , n

Since the equation of a plane is lx + my + nz = p where p is the distance from the origin,

2x – 3y + z = 5√14

Therefore, equation of the plane is 2x – 3y + z = 5√14

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