I'm a bit unclear about how you are setting up what you call your "point of interesection" equations. But you can do this:
You know that the common normal vector for the parallel planes is < 2, 2, -1 >. Pick a point in the plane 2x + 2y - z = 1 ; we can make this easy and use ( 0, 0, -1 ). Then this normal line passing through the chosen point has the vector equation
< x, y, z > = < 2t , 2t, -t - 1 > .
This line intersects the second plane for 4x + 4y - 2z = 5 , or
4 · 2t + 4 · 2t - 2 ( -t - 1 ) = 5 --> 18t + 2 = 5 ---> t = 1/6 .
The point of intersection in that plane is then ( 2 · 1/6 , 2 · 1/6 , -1/6 - 1 ) = ( 1/3 , 1/3 , -7/6 ). The distance between the two points along this line mutually perpendicular to the two planes (what is called the "perpendicular distance", the shortest distance between the planes) is given by
[tex]\sqrt{ ( \frac{1}{3} - 0 )^{2} + ( \frac{1}{3} - 0 )^{2} + ( -\frac{7}{6} - [-1] )^{2} } = \sqrt{ ( \frac{1}{3} )^{2} + ( \frac{1}{3} )^{2} + ( -\frac{1}{6} )^{2} } = \sqrt{ \frac{1}{9} + \frac{1}{9} + \frac{1}{36} }[/tex]
[tex]= \sqrt{ \frac{4 + 4 + 1}{36} } = \sqrt{ \frac{9}{36} } = \frac{1}{2} .[/tex]
I think the problem you may have made for yourself is that you didn't actually choose a point of intersection in either plane to build a normal line from.
A more general argument along these lines gives the perpendicular distance between two parallel planes ax + by + cz = d1 and ax + by + cz = d2 as [itex]D = \frac{\vert d_{1} - d_{2} \vert}{\sqrt{a^{2} + b^{2} + c^{2} } } .[/itex] For this problem, we would write
2x + 2y - z = 1 and 2x + 2y - z = 5/2 ;
The square root in the denominator gives 3 and the absolute value of the difference in the numerator is 3/2 , so the formula yields D = (3/2) / 3 = 1/2 .