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

Given: ${\mathrm{r}}_1=12, {\mathrm{r}}_2=6, {\mathrm{r}}_3=4$

$\because {\mathrm{r}}_1=12$

$\Rightarrow \frac{∆}{s-\mathrm{a}}=12$

$\Rightarrow s-\mathrm{a}=\frac{∆}{12} ...\left(1\right)$

Also,  ${\mathrm{r}}_2=6$

$\Rightarrow \frac{∆}{s-\mathrm{b}}=6$

$\Rightarrow s-\mathrm{b}=\frac{∆}{6} .....\left(2\right)$

Also, ${\mathrm{r}}_3=4$

$\Rightarrow \frac{∆}{s-\mathrm{c}}=4$

$\Rightarrow s-\mathrm{c}=\frac{∆}{4} ......\left(3\right)$

Adding $\left(1\right), \left(2\right) \& \left(3\right)$, we have

$3s-\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)=\frac{∆}{12}+\frac{∆}{6}+\frac{∆}{4}$

$\Rightarrow s=∆\frac{\left(1+2+3\right)}{12}$

$\therefore ∆=2s$

From $\left(1\right) \left(2\right) \& \left(3\right)$

Also, $∆=12\left(s-\mathrm{a}\right)=\mathrm{b}\left(s-\mathrm{b}\right)=4\left(s-\mathrm{c}\right)$

$\Rightarrow 2s=12\left(s-\mathrm{a}\right)=6\left(s-\mathrm{b}\right)=4\left(s-\mathrm{c}\right)$

$\therefore s-\mathrm{a}=\frac{2s}{12}, s-\mathrm{b}=\frac{2s}{6}, s-\mathrm{c}=\frac{2s}{4}$

$\therefore \mathrm{a}=\frac{5}{6}s, \mathrm{b}=\frac{2}{3}s, \mathrm{c}=\frac{s}{2}$

Also, $∆=\sqrt{s\left(s-\mathrm{a}\right)\left(s-\mathrm{b}\right)\left(s-\mathrm{c}\right)}$

$\Rightarrow 2s=\sqrt{s\left(\frac{2s}{12}\right)\left(\frac{2s}{6}\right)\left(\frac{2s}{4}\right)}$

$\Rightarrow 4s^2=\frac{s^4}{36}$

$\therefore s=12$

Now,

$\mathrm{a}+2\mathrm{b}+3\mathrm{c}=\frac{5}{6}s+2\times \frac{2}{3}s+3\times \frac{s}{2}$

$=\frac{5}{6}\times 12+2\times \frac{2}{3}\times 12+3\times \frac{12}{6}$

$=44$