If in a triangle ABC, r_1=2, r_2=3 and r_3=6, | vedclass

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(D) Given, $r_1=2, r_2=3$ and $r_3=6$. We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$, where $r$ is the inradius. $\frac{1

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$3$

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(D) Given, $r_1=2, r_2=3$ and $r_3=6$. We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$, where $r$ is the inradius. $\frac{1}{r} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$, so $r=1$. Also, $\Delta = \sqrt{r r_1 r_2 r_3} = \sqrt{1 	imes 2 	imes 3 	imes 6} = \sqrt{36} = 6$. Since $r_1 = \frac{\Delta}{s-a}$, we have $2 = \frac{6}{s-a}$, which implies $s-a = 3$. Also, $s = \frac{\Delta}{r} = \frac{6}{1} = 6$. Substituting $s=6$ into $s-a=3$, we get $6-a=3$, so $a=3$.

If in a $\triangle ABC$, $r_1=2$, $r_2=3$ and $r_3=6$, then $a$ equals to

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