We have $a+b+c=14$ $=>$ $(a+b+c)^2$ $=196=a^2+b^2+c^2+2(ab+bc+ca)=84+2(ab+bc+ca)$ $=>$ $ab+bc+ca=56$ $(a+b+c)^3=2744=a^3+b^3+c^3+3(a+b)(b+c)(c+a)=584+3(a+b)(b+c)(c+a)$ $=>(a+b)(b+c)(c+a)=720$ $(a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b)=a^2+b^2+c^2+3(ab+bc+ca)=84+168=252$ We have these equations: $(a+b)+(b+c)+(c+a)=28$ $(a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b)=252$ $(a+b)(b+c)(c+a)=720$ So that $a+b,b+c,c+a$ are roots of equation : $x^3-28x^2+252x-720=0$ So $a+b,b+c,c+a$ can be $6$,$10$,$12$. From here we can work out the cases and get values of $a,b,c$

Remark: you can find the equation $x^3-28x^2+252x-720=0$ using Viete, with the following formulas: ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$ and ${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(ab+bc+ca)$