Finding a point between two planes.

Question

Find a point that is midway between the two parallel planes, $x + 2y - z = -1$ and $3x + 6y - 3z = 4$.

I was able to find that the distance between the two planes is $\frac{7}{3\sqrt{6}}$, but don't know where to go from here.

Answer 1

Guide:

The first plane can be rewritten as

$$3x+6y-3z=-3$$

The second plane is

$$3x+6y-3z=4$$

Can you write out the equation of the plane that is right in the midway? Can you find one particular point on that plane?

Answer 2

Hint: $(1,2,-1)$ is perpendicular to the planes...

So, take a point on one plane, say $(0,0,1)$, and add half the distance times $(1,2,-1)$...

Get $\frac7{6\sqrt 6}\cdot(1,2,-1)+(0,0,1)=(\frac7{6\sqrt6},\frac7{3\sqrt6},-\frac7{6\sqrt6}+1)$