Explanation

(b) (i)Gravitational force on the particle placed at mid point $D$ of side $BC$ of length $a$ is
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$\overrightarrow{\mathbf{F}}\mathbf{=}{\overrightarrow{\mathbf{F}}}_{1}\mathbf{+}{\overrightarrow{\mathbf{F}}}_{2}\mathbf{+}{\overrightarrow{\mathbf{F}}}_{3}$
Here, ${\overrightarrow{\mathbf{F}}}_{2}\mathbf{= -}{\overrightarrow{\mathbf{F}}}_{3}$
$\therefore{\overrightarrow{\mathbf{F}}}_{\mathbf{\ }}\mathbf{=}{\overrightarrow{\mathbf{F}}}_{1} + 0 = {\overrightarrow{\mathbf{F}}}_{1}$
or $F = F_{1} = \frac{GMM}{\lbrack AD\rbrack^{2}} = \frac{GM^{2}}{(3a^{2}/4)} = \frac{4GM^{2}}{3a^{2}}$
(ii)gravitational force on the particle placed at the point $O$, $ie$ the intersection of three medians is
${\overrightarrow{\mathbf{F}}}_{\mathbf{\ }} = {\overrightarrow{\mathbf{F}}}_{1} + {\overrightarrow{\mathbf{F}}}_{2} + {\overrightarrow{\mathbf{F}}}_{3} = {\overrightarrow{\mathbf{0}}}_{\ }$ or $F = 0$
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Since, the resultant of ${\overrightarrow{\mathbf{F}}}_{2}$ and ${\overrightarrow{\mathbf{F}}}_{3}$ is equal and opposite to ${\overrightarrow{\mathbf{F}}}_{1}$