Ratio and proportion problem 
Question:

If $$(a+b):(b+c):(c+a)=6:7:8$$ and $a+b+c=14$, then find the value of $c$.


My solution:


*

*$$\frac{(a+b)(c+a)}{(b+c)}=\frac{(6)(8)}{7}$$$$\Rightarrow \frac{ac + a^2 + bc + ba}{b+c} = \frac{48}{7}$$$$\Rightarrow \frac{a(b+c)+a^2+bc}{b+c}=\frac{48}{7}$$$$\Rightarrow ????$$

*$$\Rightarrow a+b=6x \space\space \text{and} \space \space b+c=7x$$$$\Rightarrow b=6x-a\space\space\space\text{and}\space\space\space b=7x-c$$$$\Rightarrow \text{solving we get}\space\space x = c-a$$$$\Rightarrow ????$$

My query:
I am totally stuck on this problem. Please help. 
Thanks a lot!

$$\frac{a+b}6=\frac{b+c}7=\frac{c+a}8=\frac{-(a+b)+(b+c)+(c+a)}{-6+7+8}=\frac{2c}9$$
Similarly, each ratio is equal to $$\frac{a+b+(b+c)+(c+a)}{6+7+8}=\frac{2(a+b+c)}{21}$$
$$\implies \frac{2c}9=\frac{2(a+b+c)}{21}$$
Now, we have $a+b+c=14$
