Find the distance between the parallel planes `2x-y+3z+40` and `6x-3y+9z-3=0`.

Explanation

Clearly, we have `a_(1)/a_(2)=b_(1)/b_(2)=c_(1)/c_(2)`.
`therefore` the given planes are parallel.
Let `P(x_(1),y_(1),z_(1))` be any point on the plane `2x-y+3z+4=0`.
Then, `2x_(1)-y_(1)+3z_(1)+4=0 rArr 2x_(1)-y_(1)+3z_(1)=-4`…………………..(i)
Let p be the distance of the point `P(x_(1),y_(1),z_(1))` from the plane `6x-3y+9z-3=0`. Then,
`p=|6x_(1)-3y_(1)+9z_(1)-3|/(sqrt(6^(2)+(-3)^(2)+9^(2))=|3(2x_(1)-y_(1)+3z_(1))-3|/sqrt(36+9+81)`.
`=|3 xx (-4)-3|/sqrt(126)` [using (i)]
`=|-15|/(3sqrt(14))=15/(3sqrt(14))=5/sqrt(14)` units.
Hence, the distance between the given planes is `5/sqrt(14)` units.