Write the normal form of the equation of the plane

2x–3y + 6z + 14 = 0.

The given equation of the plane

2x–3y + 6z + 14 = 0


2x–3y + 6z = – 14 ……(i)


Now,


Dividing (i) by 7, we get,



Multiplying both sides by – 1, we get



This is the normal form of the given equation of the plane.


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