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(A) Let the mass of $A$ be $4m$ and the mass of $B$ be $3m$. Let the mass of $C$ be $M$.Let the distance of $C$ from $A$ be $r$. Then the distance of

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$\frac{2}{3}\, m$

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(A) Let the mass of $A$ be $4m$ and the mass of $B$ be $3m$. Let the mass of $C$ be $M$.Let the distance of $C$ from $A$ be $r$. Then the distance of $C$ from $B$ is $(1 - r)$.The gravitational force between $A$ and $C$ is $F_{AC} = \frac{G(4m)M}{r^2}$.The gravitational force between $B$ and $C$ is $F_{BC} = \frac{G(3m)M}{(1 - r)^2}$.According to the problem,$F_{AC} = \frac{1}{3} F_{BC}$.Substituting the expressions,we get: $\frac{G(4m)M}{r^2} = \frac{1}{3} \cdot \frac{G(3m)M}{(1 - r)^2}$.Simplifying,we get: $\frac{4}{r^2} = \frac{1}{(1 - r)^2}$.Taking the square root on both sides: $\frac{2}{r} = \frac{1}{1 - r}$.$2(1 - r) = r \Rightarrow 2 - 2r = r \Rightarrow 3r = 2 \Rightarrow r = \frac{2}{3}\, m$.

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