Two point masses A and B having masses in the ratio 4 : 3 are separated by a distance of I m. When another point mass C of mass M is placed in between A and B, the force between A and C is ${\left(\frac{1}{3}\right)}^{\mathrm{rd}}$ of the force berween B and C. Then the distance of C from A is 

Let a point mass C is placed at a distance of x m from the point mass A as shown in the figure.

Here, $\frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}} } = \frac{4}{3}$

Force between A and C is

${\mathrm{F}}_{\mathrm{AC}} = \frac{{\mathrm{GMM}}_{\mathrm{A}}}{{\mathrm{x}}^2}-------\left(\mathrm{i}\right)$

Force between B and C is

${\mathrm{F}}_{\mathrm{BC}} = \frac{{\mathrm{GMM}}_{\mathrm{B}}}{{(1-\mathrm{x})}^2}------\left(\mathrm{ii}\right)$

According to given problem

${\mathrm{F}}_{\mathrm{AC}} = \frac{1}{3}{\mathrm{F}}_{\mathrm{BC}}$

$\therefore \frac{{\mathrm{GM}}_{\mathrm{A}}\mathrm{M}}{{\mathrm{x}}^2} = \frac{1}{3}\left(\frac{{\mathrm{GM}}_{\mathrm{B}}\mathrm{M}}{{(1-\mathrm{x})}^2}\right) (\mathrm{using} (\mathrm{i}) \mathrm{and} (\mathrm{ii}\left)\right)$

$\frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{x}}^2} = \frac{{\mathrm{M}}_{\mathrm{B}}}{3{(1-\mathrm{x})}^2} \mathrm{or} \frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}}} = \frac{{\mathrm{x}}^2}{3{(1-\mathrm{x})}^2}$

$\frac{4}{3} = \frac{{\mathrm{x}}^2}{3{(1-\mathrm{x})}^2} \mathrm{or} 4 = \frac{{\mathrm{x}}^2}{{(1-\mathrm{x})}^2}$

or $2 = \frac{\mathrm{x}}{1-\mathrm{x}} \mathrm{or} 2-2\mathrm{x} = \mathrm{x}$

   3x = 2 or x = $\frac{2}{3}\mathrm{m}$