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.