The smallest positive root of the equation $\tan x - x = 0$ lies in the interval

In $\triangle ABC$,if $\cos A \cdot \cos B \cdot \cos C = \frac{1}{5}$,then $\tan A \tan B + \tan B \tan C + \tan C \tan A = $

The angles $A, B$ and $C$ of a triangle $ABC$ are in $A.P.$ and $a : b = 1 : \sqrt{3}$. If $c = 4 \text{ cm}$,then the area (in $\text{sq. cm}$) of this triangle is:

If in a triangle $ABC$,$b \cos^2 \frac{A}{2} + a \cos^2 \frac{B}{2} = \frac{3}{2} c$,then $a, b, c$ are in:

Match the items of List-$I$ with those of List-$II$ (Here $\Delta$ denotes the area of $\triangle ABC$.)

List-$I$	List-$II$
$(A)$ $\sum \cot A$	$(i)$ $\frac{(a+b+c)^2}{4\Delta}$
$(B)$ $\sum \cot \frac{A}{2}$	$(ii)$ $\frac{a^2+b^2+c^2}{4\Delta}$
$(C)$ If $\tan A : \tan B : \tan C = 1 : 2 : 3$,then $\sin A : \sin B : \sin C =$	$(iii)$ $8 : 6 : 5$
$(D)$ If $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$	$(iv)$ $12 : 5 : 13$
	$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
	$(vi)$ $4\Delta$

Then the correct match is

If the incircle of the $\Delta ABC$ touches its sides at $L, M$ and $N$ respectively,and if $x, y, z$ are the circumradii of the triangles $\Delta MIN, \Delta NIL$ and $\Delta LIM$ respectively,where $I$ is the incentre,then the product $xyz$ is equal to: