Answer
Particle 3 should be placed at the point $x=\frac{L}{1+\sqrt2}$
Work Step by Step
Take particle 3 with mass $M$, its distance to particle 1 to be $a$ and its distance to particle 2 to be $b$ (thus, $a+b=L$)
1) After particle 3 is introduced, there are 2 gravitational forces exerted on particle 1
- The gravitational force exerted by particle 2 $F_{21}$
- The gravitational force exerted by particle 3 $F_{31}$
So the magnitude of the gravitational force on particle 1 doubles when $$F_{31}+F_{21}=2F_{21}$$ $$F_{31}=F_{21}$$
2) Similarly, there are 2 gravitational forces exerted on particle 2
- The gravitational force exerted by particle 2 $F_{12}$
- The gravitational force exerted by particle 3 $F_{32}$
The magnitude of the gravitational force on particle 2 doubles when $$F_{32}+F_{12}=2F_{12}$$ $$F_{32}=F_{12}$$
We know that $F_{21}=F_{12}$. Therefore, $F_{31}=F_{32}$
As particle 1 has mass $m$ and particle 2 has mass $2m$, $$G\frac{Mm}{a^2}=G\frac{M2m}{b^2}$$ $$\frac{1}{a^2}=\frac{2}{b^2}$$ $$\frac{b}{a}=\sqrt2$$
As $a+b=L$, we can figure out that $a=\frac{L}{1+\sqrt2}$, which is the point on the x-axis particle 3 should be placed.
