The position vector of the center of mass → r cm of an asymmetric...

The position vector of center of mass $\overset{\to}{r}cm$ is given as

\overset{\to}{r}cm = {X}_{cm}\overset{^}{x} + {Y}_{cm}\overset{^}{y}

Where, ${X}_{cm} =$$x -$coordinate of center of mass and ${Y}_{cm} = y -$coordinate of center of mass

\overset{\to}{r}cm = \left(\dfrac{{m}_{1}{x}_{1} + {m}_{2}{x}_{2} + {m}_{3}{x}_{3}}{{m}_{1} + {m}_{2} + {m}_{3}}\right)\overset{^}{x} + \left(\dfrac{{m}_{1}{y}_{1} + {m}_{2}{y}_{2} + {m}_{3}{y}_{3}}{{m}_{1} + {m}_{2} + {m}_{3}}\right)\overset{^}{y}

\overset{\to}{r}cm = \left[\dfrac{\left(2m\right) \times L + \left(m\right) \times 2L + m\left(\dfrac{5L}{2}\right)}{2m + m + m}\right]\overset{^}{x} + \left[\dfrac{\left(2m\right) \times L + \left(m\right) \times \dfrac{L}{2} + m \times \left(0\right)}{2m + m + m}\right]\overset{^}{y}

\overset{\to}{r}cm = \dfrac{13L}{8}\overset{^}{x} + \dfrac{5L}{8}\overset{^}{y}