In $\triangle ABC$,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} =$

In $\triangle ABC$,if $a=13 \text{ cm}, b=14 \text{ cm}$ and $c=15 \text{ cm}$,then its circumradius $R$ is:

If $I$ is the incentre of $\triangle ABC$ and $P_1, P_2, P_3$ are respectively the radii of the circumcircles of the $\triangle IBC, \triangle ICA$ and $\triangle IAB$,then $P_1 P_2 P_3=$

$AD, BE$ and $CF$ are the perpendiculars from the vertices of a $\Delta ABC$ to the opposite sides. The ratio of the perimeter of $\Delta DEF$ to the perimeter of $\Delta ABC$ is: (where $r$ is the inradius and $R$ is the circumradius of $\Delta ABC$)

In a $\triangle ABC$, $\frac{\Delta^2}{a^2+b^2+c^2}\left(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}\right) = $

If $a, b, c$ are the sides of a $\Delta ABC$ and exradii $r_1, r_2, r_3$ are respectively $12, 6, 4$, then $a+2b+3c=$