Consider the Gaussian surface as shown in the figure.
Let the charge on the outer surface of the left-most plate be q. Thus, the charges on the plates are distributed as shown in the diagram.
The net field at point P due to all the induced charges must be zero, as it is lying inside the metal surface.
Let the surface area of the plates be A.
Electric field at point P due to the charges on plate X:
Due to charge (+Q - q) is $\frac{\text{Q}-\text{q}}{2\text{A}\in_0}$ in the right direction
Due to charge (+q) is $\frac{\text{q}}{2\text{A}\in_0}$ in the right direction
Electric field at point P due to charges on plate Y:
Due to charge (-q) is $\frac{\text{q}}{2\text{A}\in_0}$ in the right direction
Due to charge (+q) is $\frac{\text{q}}{2\text{A}\in_0}$ in the left direction
Electric field at point P due to charges on plate Z:
Due to charge (-q) is $\frac{\text{q}}{2\text{A}\in_0}$ in the right direction
Due to charge (-2Q + q) is $\frac{\text{2Q}-\text{q}}{2\text{A}\in_0}$ in the right direction
The net electric field at point P:
$\frac{\text{Q}-\text{q}}{2\text{a}\in_0}+\frac{\text{q}}{2\text{A}\in_0}-\frac{\text{q}}{2\text{A}\in_0}-\frac{\text{q}}{2\text{A}\in_0}+\frac{\text{q}}{2\text{A}\in_0}+\frac{\text{2Q}-\text{q}}{2\text{A}\in_0}=0$
$\frac{\text{Q}-\text{q}}{2\text{A}\in_0}+\frac{\text{2Q}-\text{q}}{2\text{A}\in_0}=0$
$\text{Q}-\text{q}+\text{2Q}-\text{q}=0$
$\text{3Q}-2\text{q}=0$
$\text{q}=\frac{\text{3Q}}{2}$
Thus, the charge on the outer plate of the right-most plate
$-2\text{Q}+\text{q}=-2\text{Q}+\frac{3\text{Q}}{2}=-\frac{\text{Q}}{2}$
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