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?

The sides $a, b, c$ of a triangle satisfy the relations $c^2=2ab$ and $a^2+c^2=3b^2$. Then the measure of $\angle BAC$,in degrees,is

The greatest angle of the triangle whose sides are $x^2+x+1$,$2x+1$,and $x^2-1$ is (in $^{\circ}$)

In a non-right-angled triangle $\triangle PQR$, let $p, q, r$ denote the lengths of the sides opposite to the angles at $P, Q, R$ respectively. The median from $R$ meets the side $PQ$ at $S$, the perpendicular from $P$ meets the side $QR$ at $E$, and $RS$ and $PE$ intersect at $O$. If $p=\sqrt{3}, q=1$, and the radius of the circumcircle of the $\triangle PQR$ equals $1$, then which of the following options is/are correct?
$(1)$ Area of $\triangle SOE = \frac{\sqrt{3}}{48}$
$(2)$ Radius of incircle of $\triangle PQR = \frac{\sqrt{3}}{2}(2-\sqrt{3})$
$(3)$ Length of $RS = \frac{\sqrt{7}}{2}$
$(4)$ Length of $OE = \frac{1}{6}$

Let $S=\{x \in(-\pi, \pi) \mid x \neq 0, \pm \frac{\pi}{2}\}$. The sum of all distinct solutions of the equation $\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$ in the set $S$ is equal to

In a triangle $ABC$,$\left(\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}\right)^2 \leq$