The distance between the planes $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$ is
A) $\dfrac{{\sqrt 7 }}{{2\sqrt 2 }}$
B) $\dfrac{7}{2}$
C) $\dfrac{{\sqrt 7 }}{2}$
D) $\dfrac{7}{{2\sqrt 2 }}$

Hint:
Here, we are asked to find the distance between the planes $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$.
Firstly, compare the equations of the planes $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$ with $Ax + By + Cz + {D_n} = 0$ and find $A,B,C,{D_1},{D_2}$.
Thus, find the distance between two planes using the formula $d = \dfrac{{\left| {{D_1} - {D_2}} \right|}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}$.

Complete step by step solution:
The given planes have the equations $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$ .
Now, the equation of second plane can be written as \[2\left( {x + 2y + 3z + \dfrac{7}{2}} \right) = 0 \Rightarrow x + 2y + 3z + \dfrac{7}{2} = 0\] .
On comparing the equations of planes with $Ax + By + Cz + {D_n} = 0$ , we get $A = 1,B = 2,C = 3,{D_1} = 7,{D_2} = \dfrac{7}{2}$ .
Now, the distance between two parallel planes is given by $d = \dfrac{{\left| {{D_1} - {D_2}} \right|}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}$
Thus, $d = \dfrac{{\left| {7 - \dfrac{7}{2}} \right|}}{{\sqrt {{1^2} + {2^2} + {3^2}} }}$
 $
   = \dfrac{{\dfrac{7}{2}}}{{\sqrt {1 + 4 + 9} }} \\
   = \dfrac{7}{{2\sqrt {14} }} \\
   = \dfrac{{\sqrt 7 \times \sqrt 7 }}{{2\sqrt 2 \times \sqrt 7 }} \\
   = \dfrac{{\sqrt 7 }}{{2\sqrt 2 }} \\
 $

So, option (A) is correct.

Note:
This question can be solved by the help of some basic concepts of the distance between planes from the chapter three dimensional geometry.
Here, student may go wrong by finding the distance between the planes as $d = \left| {{D_1} - {D_2}} \right|$ , which cannot be the answer.
Thus, when we are solving these types of question take into consideration that find distance between planes using the formula $d = \dfrac{{\left| {{D_1} - {D_2}} \right|}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}$.