In $\triangle ABC$,if $\tan A + \tan B + \tan C = 6$ and $\tan A \cdot \tan B = 2$,then $\tan C = \dots$

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

The number of solutions to $\sin x = \frac{6}{x}$ with $0 \leq x \leq 12 \pi$ is

If the angles $A, B$ and $C$ of a triangle are in an Arithmetic Progression and if $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively,then the value of the expression $\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A$ is

The angle bisectors $BD$ and $CE$ of a $\triangle ABC$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively. Then,the ratio in which $I$ divides the angle bisector through $A$ is

In a triangle,if $r_1 = 2r_2 = 3r_3$,then $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is equal to