In $∆\mathrm{ABC},$ if ${\mathrm{r}}_1=2{\mathrm{r}}_2=3{\mathrm{r}}_3$ and D is the mid point of BC then cos $\angle \mathrm{ADC}=$

$\begin{array}{l}{r}_1=2r_1=3r_3\\ \Rightarrow \frac{\mathrm{\Delta }}{s-a}=\frac{2\mathrm{\Delta }}{s-b}=\frac{3\mathrm{\Delta }}{s-c}=\frac{\mathrm{\Delta }}{k}\text{ (say) }\\ \Rightarrow s-a=k,s-b=2k,s-c=3k\\ \Rightarrow 3s-(a+b+c)=6k\Rightarrow s=6k\\ \Rightarrow \frac{a}{5}=\frac{b}{4}=\frac{c}{3}=k\\ \text{ So, }{a}^2=b^2+c^2\\ \Delta ABC is right angled \Delta , \angle A = {90}^0 and D is midpoint of BC, \\ AD = DC (radius of circumicrcle)\\ \begin{array}{l}\Rightarrow \mathrm{\angle }DAC=C\Rightarrow \mathrm{\angle }ADC={180}^{\circ }-2C\\ \Rightarrow \cos \mathrm{\angle }ADC\\ =\cos \left({180}^{\circ }-2C\right)\\ =-\cos 2C\\ -\left(2{\cos }^{2}C-1\right)\\ =1-2{\cos }^{2}C\\ =1-2\times \frac{16}{25}=\frac{-7}{25}\end{array}\end{array}$