${\text{In a △ABC, if r=1,R=3,s=5, then the value of a}}^2+b^2+c^2 is$

$\begin{array}{l}\text{Given} r=1,s=5 △=rs=5\\ \text{Also} s=5 \Rightarrow a+b+c=10\text{}\\ abc=4R△\text{}\\ abc=4\times 3\times 5=60\text{}\\ \text{now,} {△}^2=s\left(s-a\right)\left(s-b\right)\left(s-c\right)\\ \text{}25=5\left(5-a\right)\left(5-b\right)\left(5-c\right)\\ \text{}5=125-25\left(a+b+c\right)+5\left(ab+bc+ca\right)-abc\\ \text{}5=125-25x10+5\left(ab+bc+ca\right)-60\text{}\\ ab+bc+ca=38\text{}\\ \text{hence}, a^2+b^2+c^2={\left(a+b+c\right)}^2-2\left(38\right)=24\end{array}$