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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