Distance between two parallel planes 2x+y+2z=8 and 4x+2y+4z+5=0 is

Given equation of planes are 2x+y+2z=8 and 4x+2y+4z+5=0
⇒ 4x+2y+4z=16
⇒ 4x+2y+4z=-5
The distance between two parallel planes ax+by+cz+d1=0 and ax+by+cz+d2=0 is given by
dmin= |d1-d2|/√(a²+b²+c²)
Here, d1=-5, d2=16, a=4, b=2, c=4
dmin=|16-(-5)|/√(4²+2²+4²)=21/√36=21/6=7/2=3.5
However, the given options do not include 3.5 or 7/2.
Let's use the equations as 2x+y+2z-8=0 and 2x+y+2z+5/2=0
dmin=| -8 - (5/2)| / √(2²+1²+2²) = |-21/2| / 3 = 21/6 = 7/2 = 3.5
Let's reconsider the given equations: 2x+y+2z=8 and 4x+2y+4z+5=0. We can rewrite the second equation as 2(2x+y+2z) = -5, or 2x+y+2z = -5/2.
The distance between the planes is given by:
d = |d1 - d2| / sqrt(a^2 + b^2 + c^2)
where d1 = 8, d2 = -5/2, a = 2, b = 1, c = 2
d = |8 - (-5/2)| / sqrt(2^2 + 1^2 + 2^2) = |21/2| / sqrt(9) = (21/2) / 3 = 7/2 = 3.5
Since 3.5 is not among the options, there might be a mistake in the question or options provided. The correct distance is 3.5 or 7/2.