In a triangle ABC, if r_1=2 r_2=3 r_3, then f | vedclass

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(D) Given $r_1=2 r_2=3 r_3$.Using the formula $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, $r_3 = \frac{\Delta}{s-c}$, we have:$\frac{\Delt

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$\frac{191}{60}$

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(D) Given $r_1=2 r_2=3 r_3$.Using the formula $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, $r_3 = \frac{\Delta}{s-c}$, we have:$\frac{\Delta}{s-a} = \frac{2\Delta}{s-b} = \frac{3\Delta}{s-c}$From $\frac{1}{s-a} = \frac{2}{s-b}$, we get $s-b = 2s-2a \Rightarrow s = 2a-b$.From $\frac{1}{s-a} = \frac{3}{s-c}$, we get $s-c = 3s-3a \Rightarrow 2s = 3a-c$.Substituting $s = \frac{a+b+c}{2}$ into these equations:$a+b+c = 4a-2b \Rightarrow 3a-3b = c$.$a+b+c = 3a-c \Rightarrow 2a-b = 2c$.Solving for ratios:From $3a-3b = c$ and $2a-b = 2c$, we get $2(3a-3b) = 2a-b$ $\Rightarrow 6a-6b = 2a-b$ $\Rightarrow 4a = 5b$ $\Rightarrow \frac{a}{b} = \frac{5}{4}$.Then $c = 3a-3b = 3a - 3(\frac{4a}{5}) = 3a - \frac{12a}{5} = \frac{3a}{5} \Rightarrow \frac{c}{a} = \frac{3}{5}$.Since $\frac{a}{b} = \frac{5}{4}$ and $\frac{c}{a} = \frac{3}{5}$, then $\frac{b}{c} = \frac{b}{a} 	imes \frac{a}{c} = \frac{4}{5} 	imes \frac{5}{3} = \frac{4}{3}$.Finally, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} = \frac{5}{4} + \frac{4}{3} + \frac{3}{5} = \frac{75+80+36}{60} = \frac{191}{60}$.

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

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