Find the distance between the planes `x+2y+3z+7=0` and `2x+4y+6z+7=0`.

Answer

`7/sqrt(56)` units.

Explanation

Let `P(x_(1),y_(1)z_(1))` be any point on the plane `x+2y+3z+7=0`.
Then, `x_(1)+2y_(1)+3z_(1)=-7`.
`therefore p=(|2x_(1)+4y_(1)+6z_(1)+7|)/(sqrt(2^(2)+4^(2)+6^(2))=(|2(x_(1)+2y_(1)+3z_(1))+7|)/sqrt(56)`………….(i)
`=(|2 xx (-7) +7|)/sqrt(56)`.............(i)
`=(|2xx(-7)+7|)/sqrt(56)=7/sqrt(56)` units.