Since, $|r-b|+|r-c|=4$ is equation of ellipsoid in plane $r,a=5$ or $x+y+z=5$.

$\therefore$ Distance between foci $=|b-c|$

$=|3\hat{\mathrm{i}}-\hat{\mathrm{j}}-2\hat{\mathrm{k}}|=\sqrt{14}=2ae$

and $2a=4$

$\therefore e=\frac{\sqrt{14}}{4}\Rightarrow e^2=1-\frac{b^2}{a^2}$

$\Rightarrow \frac{14}{16}=1-\frac{b^2}{4}\Rightarrow \frac{b^2}{4}=\frac{2}{16}$

$b^2=\frac{1}{2}$

$\therefore$ Area of ellipsoid 

$=\pi ab=\pi \left(2\right)\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2}\pi$

And from the given options the closest integer to $\sqrt{2}\pi$ is $4$.