Think about a triangle

Rewriting the equations,

$\left(a+b\right)^2-ab=25$

$\left(b+c\right)^2-bc=49$

$\left(a+c\right)^2-ac=64$

Now subtracting $1$ from $2$ , $2$ from $3$ , and $1$ from $3$ ,

$\left(a+b+c\right)\left(c-a\right)=24$

$\left(a+b+c\right)\left(a-b\right)=15$

$\left(a+b+c\right)\left(c-b\right)=39$

Hence, we have:

$\frac{24}{c-a}=\frac{15}{a-b}=\frac{39}{c-b} \implies \frac{8}{c-a}=\frac{5}{a-b}=\frac{13}{c-b} \implies \frac{8}{c-a}=\frac{5}{a-b} \implies \frac{64}{a^2+c^2-2ac}=\frac{25}{a^2+b^2-2ab} \implies \frac{64}{a^2+c^2+ac-3ac}=\frac{25}{a^2+b^2+ab-3ab} \implies \frac{64}{64-3ac}=\frac{25}{25-3ab}$

Cross multipltying, we have $64b=25c \implies c=\frac{64b}{25}$

Substituting $\frac{64b}{25}$ in (2), we have:

$\left(b+\frac{64b}{25}\right)^2-\frac{64b^2}{25}=49 \implies b=\frac{25}{\sqrt{129}}$ (b is a positive real)

$c=\frac{64b}{25}=\frac{64}{25}\cdot \frac{25}{\sqrt{129}}=\frac{64}{\sqrt{129}}$

Substituting $c=\frac{64}{\sqrt{129}}$ in (3), we have $a=\frac{40}{\sqrt{129}}$ (a is a positive real)

Hence, $\left(a+b+c\right)^2=\left(\frac{40}{\sqrt{129}}+\frac{25}{\sqrt{129}}+\frac{64}{\sqrt{129}}\right)^2=\left(\frac{129}{\sqrt{129}}\right)^2=\boxed{129}$

Consider a triangle $ABC$ with sides $AB=5,\,BC=7,\,AC=8$ , and let $P$ be the Steiner point inside the triangle which subtends an angle of $120^\circ$ with each pair of the points $A,B,C$ . Note that $P$ can be constructed by constructing the equilateral triangle $ABX$ , in which case $P$ is the point of intersection (other than $X$ ) of the line $CX$ with the circumcircle of $ABX$ . Suppose that $a = AP$ , $b = BP$ and $c = CP$ . Applying the Cosine Rule to each of the triangles $ABP,\,BCP,\,ACP$ shows that $a,b,c$ satisfy the desired identities.

Rotate the triangle $ABP$ through $60^\circ$ clockwise about the point $A$ to form the triangle $AXP'$ . Note that $P'$ lies on the line $XC$ . Then $APP'$ is equilateral, and it is clear that $a + b + c \; = \; AP + BP + CP \; = \; PP' + P'X + CP \;= \; CX$

Set up a coordinate system so that $A$ has coordinates $(0,0)$ and $B$ has coordinates $(5,0)$ . Then $X$ has coordinates $(\tfrac52,-\tfrac52\sqrt{3})$ . Using the Cosine rule on the triangle $ABC$ , we see that $\angle CAB = 60^\circ$ , and so $C$ has coordinates $(4,4\sqrt{3})$ . Thus $(a + b + c)^2 \; = \; CX^2 \; = \; \left(\tfrac32\right)^2 + \left(\tfrac{13}{2}\sqrt{3}\right)^2 \; = \; \boxed{129} \;.$ Alternatively, we could apply the Cosine Rule to the triangle $CAX$ , with $\angle CAX = 120^\circ$ , to show that $CX^2 = 129$ .