(a) We can see from the diagram that A is normal to the x-y plane in the positive z-direction and B is parallel to the z-axis.

$\vec{A}=50 \times 10^{-4} \hat{k}$

$\vec{B}=0.3 \hat{k}$

Accordingly,

$\vec{\tau}=12 \times\left(-50 \times 10^{-4}\right) \hat{k} \times 0.3 \hat{k}$

$=0$

Hence, the torque is zero. The force is also zero.

(b) We can see from the diagram that A is normal to the x-y plane in the negative z-direction and B is parallel to the z-axis. The angle formed by A and B is $\theta=180^{\circ}$. Therefore,

$\vec{A}=-50 \times 10^{-4} \hat{k}$

$\vec{B}=0.3 \hat{k}$

$\vec{\tau}=12 \times\left(-50 \times 10^{-4}\right) \hat{k} \times 0.3 \hat{k}$

$=0$

Therefore, torque and force is zero.

Case (f) corresponds to unstable, and case (e) corresponds to stable equilibrium.