In a triangle $ABC$,if $(r_1-r_3)(r_1-r_2)-2r_2r_3=0$,then $a^2-b^2=$

In $\triangle ABC$,if $A = 60^{\circ}$ and $B = 105^{\circ}$,then find the value of $\frac{2R^2(b-c) \sin A \sin B \sin C}{(b+c)(s-a \cos C - c \cos A)(s-a \cos B - b \cos A)}$.

For a $\Delta ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides $a, b, c$ are in:

If in a triangle $ABC$,$\cos A \cos B + \sin A \sin B \sin C = 1$,then $a : b : c =$

Observe the following statements:
$(I)$ In $\triangle ABC$,$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC$,$\cot \frac{A}{2} = \frac{b+c}{a} \implies B = 90^{\circ}$
Which of the following is correct?

In a $\triangle ABC$,$\cos \left(\frac{B+2C+3A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$ is equal to