Answer
(a) C, (b) A, (c) B
Work Step by Step
1) The translational forces:
We have $\sum F=ma$
In A, $\sum F=4F$, so $a\ne0$
In B, $\sum F=2F-F-F=0$, so $a=0$
In C, $\sum F=F+F-2F=0$, so $a=0$
2) The rotational torques:
The axis of rotation is at the center, so in A and B, the force $\vec{2F}$ does not create any torques because it passes the center of the sheets.
Take the lever arm for each force to be $L$, since $\sum \tau=I\alpha$; we have
In A, $\sum\tau=FL-FL=0$, so $\alpha=0$
In B, $\sum\tau=FL-FL=0$, so $\alpha=0$
In C, $\sum\tau=FL-FL-2FL=-2FL$, so $\alpha\ne0$
Therefore, in A, $a\ne0$ but $\alpha=0$ -> (b)
In B, $a=\alpha=0$ -> (c)
In C, $a=0$ but $\alpha\ne0$ -> (a)
