For triangle sides $a,b,c\in\Bbb N$, find semiperimeter $s$ if $\frac1{36}(s-a)^2+\frac19(s-b)^2+\frac14(s-c)^2=\frac1{49}s^2$ and $HCF(a,b,c)=1$ 
Let $a,b,c$ be the lengths of sides opposite to vertex $A$, $B$, and $C$ respectively in triangle $ABC$ . If
  $$\frac{(s-a)^2}{36}+\frac{(s-b)^2}{9}+\frac{(s-c)^2}{4}=\frac{s^2}{49}$$ 
  and $a,b,c \in \Bbb{N}$, and $\operatorname{HCF}(a, b, c)=1$. Then find the value of the semi-perimeter(s).

This is some a problem which involves some calculation. If we do some calculation with denominator, we get $36+9+4=49$, but I'm not getting the required result. 

We have the case of the equality in the inequality of Cauchy-Schwarz:
$$
s^2
=\left(
\frac{(s-a)^2}{6^2}+\frac{(s-b)^2}{3^2}+\frac{(s-c)^2}{2^2}
\right)
\left(6^2+3^2+2^2\right)
\ge
\Big(\ (s-a)+(s-b)+(s-c)\ \Big)^2
=s^2\ .
$$
So the two vectors $\Big(\ \frac 16(s-a),\ \frac 13(s-b),\ \frac 12(s-c)\ \Big)$ and $(6,3,2)$ are linearly dependent. We get for some scalar $k$, $s-a=36k$, $s-b=9k$, $s-c=4k$, so $s =(s-a)+(s-b)+(s-c)=49k$. We determine 
$$
\begin{aligned}
a &=s-(s-a)=49k-36k=13k\ ,\\
b &=s-(s-b)=49k-9k=40k\ ,\\
c &=s-(s-c)=49k-4k=45k\ ,
\end{aligned}
$$
and we have to take $k=1$ now.
