A planar coil of area 7 ${\mathrm{m}}^2$ carrying an anti-clockwise current 2 A is placed in an extemal magnetic field $\vec{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})$, such that the normal to the plane is along the line$(3\hat{i}-5\hat{j}+4\hat{k})$. Select correct statements from the following .  ( Consider Normal of the coil and Magnetic moment vectors to be in  the same direction ) 

Area $A=7m^2,$ Normal $\hat{n}=\frac{1}{\sqrt{50}}(3\hat{i}-5\hat{j}+4\hat{k})$
$\simeq \frac{1}{7}(3\hat{i}-5\hat{j}+4\hat{k})$
$\therefore$ area $\vec{A}=(3\hat{i}-5\hat{j}+4\hat{k})m^2$
Magnetic field $\vec{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})T$
Current $I=2A$
$\therefore$ magnetic moment $\vec{M}=I\vec{A}=(6\hat{i}-10\hat{j}+8\hat{k})A{m}^2$
$|\vec{M}|\simeq 14A{m}^2(2\times 7=14)$
Potential energy U$=-\vec{M}.\vec{B}$
$=-[1.2-2-2.4]=3.2J$
Let $\theta$ be the angle between $\hat{n}$ and $\vec{B}$(i.e.,)between $\vec{M}$ and $\vec{B}.$
Now,U$=-MBcos\theta .$
$\therefore cos\theta =-\frac{U}{MB}=-\frac{3.2}{14\times 0.4}(\because \sqrt{0.17}\simeq 0.4)$
$=-\frac{4}{7}=-0.57$
$\theta =\pi -co{s}^{-1}(0.57)$