System of Particles on a wedge!

Let centre of mass of the system be denoted as $r_{cm}$

Initially

$(r_{cm})_{i}=\frac{m\times 0+2mL+mL}{4m}$

when the $2m$ reaches bottom of wedge , let the distance moved by centre of mass be $x$ and this new position be denoted as $(r_{cm})_ {f}$ . we have to basically find $x$

$(r_{cm})_ {f}=\frac{m(L-x)+2m(2L-x)+m(L-x)}{4m}$

Since no external force acts on the system horizontally so centre of mass remains at rest so

$(r_{cm})_{i}=(r_{cm})_{f}$

solve for $x$ , we get

$\huge{\boxed{x=\frac{3L}{4}}}$

We can solve it in one more way although with a similar concept as used by Tanishq Varshney .
If the Wedge moves backward by a distance of x then both the point masses will move by (L-x) in forward direction. As the center of mass doesnot move in the horizontal direction, Therefore it can be written that
${ \Delta x }_{ cm }={ (m }_{ 1 }{ \Delta x }_{ 1 }+{ m }_{ 2 }{ \Delta x }_{ 2 }+{ m }_{ 3 }{ \Delta x }_{ 3 })/{ m }_{ total }=0$
=> 2m(L-x)+m(L-x)-mx=0
=>x=3L/4