Answer :

Formula : Plane = r . (n) = d


Where r = any random point


n = normal vector of plane


d = distance of plane from origin


If two planes are parallel , then their normal vectors are same.


Therefore ,


Parallel Plane 2x – 3y + 7z + 13 = 0


Normal vector = (2i - 3j + 7k)


Normal vector of required plane = (2i - 3j + 7k)


Equation of required plane r . (2i - 3j + 7k) = d


In cartesian form 2x - 3y + 7z = d


Plane passes through point (0,0,0) therefore it will satisfy it.


2(0) - (0) + 3(0) = d


d = 0


Equation of required plane 2x - 3y + 7z = 0


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