(a) Let us detail each case separately,


Dipole moment $(\vec{M})$ is along $+\mathrm{x}$ direction
$$
\begin{aligned}
&\vec{M}=10 \times 5 \times 10^{-4} \times 12 \hat{i}=0.06 \hat{i} \\
&\vec{B}=3000 \times 10^{-4} \hat{k}
\end{aligned}
$$
Torque $\vec{C}=\vec{M} \times \vec{B}=-1.8 \times 10^{-2} \hat{J}$ N.m
Torque is $1.8 \times 10^{-2} \mathrm{~N}-\mathrm{m}$ along $-\mathrm{y}$ direction, Net force or the coil is zero, coil is not in equilibrium.
(b) Dipole moment is along $+\mathrm{x}$ direction
$$
\vec{M}=0.06 \hat{i}, \vec{B}=0.3 \hat{k}
$$

Torque $\vec{\tau}=\vec{M} \times \vec{B}$
So, torque is $1.8 \times 10^{-2} \mathrm{~N}-\mathrm{m}$ along $-y$ direction.
Net force on the coil is zero, coil is not in equilibrium.
(c) Dipole moment is along $-y$ direction
$\vec{M}=-0.06 \hat{j}, \hat{B}=0.3 \hat{k}$
Torque $\vec{\tau}=\vec{M} \times \vec{B}=-1.8 \times 10^{-2} \hat{i} \mathrm{~N}-\mathrm{m}$.
So, the torque is $1.8 \times 10^{-2} \mathrm{Nm}$ along $-x$ direction.
Net force on the coil is zero, coil is not in equilibrium.
(d) Dipole moment is at an angle $150^{\circ}$ with the $+x$ direction.

$$
=1.8 \times 10^{-2} \sin \frac{\pi}{2}=1.8 \times 10^{-2} \text { N.m }
$$
At an angle $240^{\circ}$ with the $+x$ direction, net force on the coil is zero, coil is not in equilibrium.
(e) Dipole moment is along $+\mathrm{z}$ direction.

Torque $\vec{\tau}=\vec{M} \times \vec{B}=1.8 \times 10^{-2} \hat{k} \times \hat{k}=0$
Potential energy $U=\vec{M} \cdot \vec{B}$
$$
U=-1.8 \times 10^{-2}(\hat{k} \hat{k})=-1.8 \times 10^{-2} \mathrm{~J}
$$
negative sign shows equilibrium is stable.
(f) Dipole moment is along $-\mathrm{z}$ direction

Torque $\vec{\tau}=\vec{M} \times \vec{B}=1.8 \times 10^{-2}(-\hat{k} \cdot \hat{k})=0$
Potential energy, $U=-\vec{M} \cdot \vec{B}=+1.8 \times 10^{-2} \mathrm{~J}$ Positive energy so, equilibrium is unstable.