Answer
$D=\frac{|d_1-d_2|}{\sqrt {a^2+b^2+c^2}}$
Work Step by Step
$n=\lt a,b,c\gt$
$a(x_1)+b(0)+c(0)+d_1=0$
$P=(\frac{-d_1}{a},0,0)$
$a(x_2)+b(0)+c(0)+d_2=0$
$Q=(\frac{-d_2}{a},0,0)$
Take a vector $B$ for the line between points $P$ and $Q$.
$B=PQ=(\frac{-d_1}{a},0,0) \cdot (\frac{-d_2}{a},0,0)$
$=(\frac{d_1-d_2}{a},0,0)$
$n.B=\lt d_1-d_2,0,0\gt$
$|n|=\sqrt {a^2+b^2+c^2}$
Therefore, the distance formula between two parallel planes is given as:
$D=\frac{|n.B|}{|n|}=\frac{|d_1-d_2|}{\sqrt {a^2+b^2+c^2}}$
