The ratio in which the line $3 x+4 y+2=0$ divides the distance between the lines $3 x+4 y+5=0$ and $3 x+4 y-5=0$ is

The given lines are parallel, represented by the equations $3x + 4y + 5 = 0$, $3x + 4y + 2 = 0$, and $3x + 4y - 5 = 0$.

Let the lines be $L_1: 3x + 4y + 5 = 0$, $L_2: 3x + 4y + 2 = 0$, and $L_3: 3x + 4y - 5 = 0$.

The distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \dfrac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.

The distance $d_1$ between $L_1$ and $L_2$ is $d_1 = \dfrac{|5 - 2|}{\sqrt{3^2 + 4^2}} = \dfrac{3}{5}$.

The distance $d_2$ between $L_2$ and $L_3$ is $d_2 = \dfrac{|2 - (-5)|}{\sqrt{3^2 + 4^2}} = \dfrac{7}{5}$.

The ratio in which $L_2$ divides the distance between $L_1$ and $L_3$ is $d_1 : d_2 = \dfrac{3}{5} : \dfrac{7}{5} = 3 : 7$.

Answer: 3 : 7