In $\mathrm{\Delta ABC},{\sin }^4\mathrm{A}+{\sin }^4\mathrm{B}+{\sin }^4\mathrm{C}={\sin }^2{\mathrm{Bsin}}^2\mathrm{C}+2{\sin }^2{\mathrm{Csin}}^2\mathrm{A}+2{\sin }^2{\mathrm{Asin}}^2\mathrm{B}$ Then A =

$\begin{array}{l}From sine rule\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}=2R\\ Given that {\sin }^4\mathrm{A}+{\sin }^4\mathrm{B}+{\sin }^4\mathrm{C}={\sin }^2{\mathrm{Bsin}}^2\mathrm{C}+2{\sin }^2{\mathrm{Csin}}^2\mathrm{A}+2{\sin }^2{\mathrm{Asin}}^2\mathrm{B}\end{array}$
$\begin{array}{l}{\mathrm{a}}^4+{\mathrm{b}}^4+{\mathrm{c}}^4={\mathrm{b}}^2{\mathrm{c}}^2+2{\mathrm{c}}^2{\mathrm{a}}^2+2{\mathrm{a}}^2{\mathrm{b}}^2\\ {\mathrm{a}}^4+{\mathrm{b}}^4+{\mathrm{c}}^4-2{\mathrm{a}}^2{\mathrm{b}}^2-2{\mathrm{c}}^2{\mathrm{a}}^2+2{\mathrm{b}}^2{\mathrm{c}}^2=3{\mathrm{b}}^2{\mathrm{c}}^2\\ {\left({\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2\right)}^2=3{\mathrm{b}}^2{\mathrm{c}}^2\\ (2\mathrm{bccos}\mathrm{A}{)}^2=3{\mathrm{b}}^2{\mathrm{c}}^2(\because \mathrm{from} \mathrm{cosine} \mathrm{rule})\\ 4{\mathrm{b}}^2{\mathrm{c}}^2{\cos }^2\mathrm{A}=3{\mathrm{b}}^2{\mathrm{c}}^2\\ {\cos }^2\mathrm{A}=\frac{3}{4}\\ \cos \mathrm{A}=\pm \frac{\sqrt{3}}{2}\Rightarrow \mathrm{A}=\frac{\mathrm{\pi }}{6}\text{ and }\frac{5\mathrm{\pi }}{6}\end{array}$