The product of the distances of the incentre $I$ from the angular points $A, B, C$ of a $\Delta ABC$ is:

In a $\Delta ABC,$ let $\angle C = \frac{\pi}{2}.$ If $r$ and $R$ are the inradius and the circumradius respectively of the triangle,then $2(r + R)$ is equal to

In a $\triangle ABC$,if $\sin^2 B = \sin A$ and $2 \cos^2 A = 3 \cos^2 B$,then the triangle is

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

Consider the following statements.
$I$. In $\triangle ABC$,if $c=6$ and $\cos C=-\frac{11}{25}$,then $R=\frac{25}{2\sqrt{14}}$.
$II$. In $\triangle ABC$,if $a=3, b=4, c=6$,then $\triangle ABC$ is an acute-angled triangle.
Which of the above statements is/are true?

Consider a triangle $PQR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$,respectively. Then which of the following statements is (are) $TRUE$?
$(A)$ $\cos P \geq 1-\frac{p^2}{2qr}$
$(B)$ $\cos R \geq \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$
$(C)$ $\frac{q+r}{p} < 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$
$(D)$ If $p < q$ and $p < r$,then $\cos Q > \frac{p}{r}$ and $\cos R > \frac{p}{q}$