Answer
$x+2y-2z-7=0$ ; $x+2y-2z+5=0$
Work Step by Step
$x+2y-2z=1$
$x+2y-2z-1=0$
$n= \lt 1,2, -2\gt $
Distance formula between two parallel planes is given as:
$D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$
$2=\frac{|x+2y-2z-1|}{\sqrt {1^2+2^2+(-2)^2}}$
$6=|x+2y-2z-1|$
First plane is:
$6=x+2y-2z-1$ $\implies$ $x+2y-2z-7=0$
Second plane is:
$6=-(x+2y-2z-1)$ $\implies$ $x+2y-2z+5=0$
Therefore, the distance formula between two parallel planes is given as:
$D=\frac{|d_1-d_2|}{\sqrt {a^2+b^2+c^2}}$
$D=\frac{|-1-(-7)|}{3}$
$D=\frac{6}{3}=2$
