Describing the nature of the intersection

Question

Describe the nature of the intersection between the planes with equations $$3x-2y+2z=4$$ $$-3x+2y-2z=9$$$$6x-4y+4z=8$$

Does anyone know how to solve this or how to start?

Answer 1

After rewriting a bit, we have:

\begin{align} 3x-2y+2z&=4 \\3x-2y+2z&=-9\\ 2(3x-2y+2z)&=2\cdot 4 \end{align} The first and second equations can never be equal (since they are the same on the LHS, but differs on the RHS), and the third is just the first multiplied by $2$, so they are the same.

Hence the intersection is empty, meaning there are no points that lie in all three planes at the same time.

Answer 2

The second equation is a scalar multiple (-1) of the first equation on the LHS, but have different constants on the RHS. This means that the planes described by these equations are parallel, and hence have no intersection.

The third equation is exactly a scalar multiple (2) of the first equation, meaning that the first and third equations form the same plane.

Therefore, there are no intersections.