Calculate the distance between the two planes:
$$\pi:x+3y-\sqrt{6}z-4=0$$
$$\pi':x+3y-\sqrt{6}z+1=0$$
See development and solution
Development:
To calculate the distance between the planes $$\pi$$ and $$\pi'$$ we can apply: $$$ \text{d}(\pi,\pi') = \dfrac{|D-D|}{\sqrt{A^2+B^2+C^2}}= \dfrac{|-4-1|}{\sqrt{1+9+6}}=\dfrac{5}{4}$$$
Solution:
$$\text{d}(\pi,\pi') = \dfrac{5}{4}$$