Thank you for reporting, we will resolve it shortly
Q.
Distance between the parallel planes 2x - y + 3z + 4 = 0 and 6x - 3y + 9z - 3 = 0 is:
Three Dimensional Geometry
Solution:
Note: The perpendicular distance from the origin to the plane ax + by + cz + d = 0 is
given by $\frac{d}{\sqrt{a^2 + b^2 + c^2}}$
Thus the perpendicular distance from origin to the plane 2x - y + 3z + 4 = 0 is
$P_1 = \frac{4}{\sqrt{14}}$
Similarly, $P_2 = \frac{-3}{3 \sqrt{14}} = \frac{-1}{\sqrt{14}}$
Thus the distance between the two parallel planes
$ = | P_2 - P_1| = \frac{5}{\sqrt{14}}$.