Tricky system

Let's rewrite the equations:

$a^2+b^2-2ab \cos 150°=727$

$a^2+c^2-2ac \cos 150°=829$

$b^2+c^2-2bc \cos 60°=39$

Clearly, by Law of Cosines, we can form three triangles, but also, since the three angles add up to $360°$ we can form a big triangle $\triangle ABC$ with an interior point $P$ such that $\angle APB=150°$ , $\angle APC=150°$ , $\angle BPC=60°$ , $PA=a$ , $PB=b$ and $PC=c$ . We know that $AB=\sqrt{727}$ , $AC=\sqrt{829}$ and $BC=\sqrt{39}$ . Using Heron's formula, we find the area of $\triangle ABC$ :

$[ABC]=\dfrac{191\sqrt{3}}{4}$

Also, the area of $\triangle ABC$ can be calculated adding the areas of $\triangle APB$ , $\triangle APC$ and $\triangle BPC$ :

$[ABC]=\dfrac{ab \sin 150°}{2}+\dfrac{ac \sin 150°}{2}+\dfrac{bc \sin 60°}{2}=\dfrac{ab+ac+\sqrt{3}bc}{4}$

Hence, $ab+ac+\sqrt{3}bc=191\sqrt{3}$ .

Now, multiply the third equation by $-3$ :

$-3b^2-3c^2+3bc=-117$

Sum it with the other two:

$2(a^2-b^2-c^2)+\sqrt{3}(ab+ac+\sqrt{3}bc)=1439$

And replace the fourth equation:

$2(a^2-b^2-c^2)+\sqrt{3}(191\sqrt{3})=1439 \Longrightarrow a^2-b^2-c^2=433$

Next, sum the third equation with the fifth equation:

$a^2-bc=433+39 \Longrightarrow a^2=bc+472$

Replace it to the first and second equation:

$b^2+bc+\sqrt{3}ab+472=727$

$c^2+bc+\sqrt{3}ac+472=829$

Sum and subtract them to obtain, respectively:

$(b+c)(\sqrt{3}a+b+c)=612$

$(b-c)(\sqrt{3}a+b+c)=-102$

Divide them:

$\dfrac{b+c}{b-c}=\dfrac{612}{-102} \Longrightarrow c=\dfrac{7b}{5}$

Substitute it in the fifth equation and in the sixth equation:

$a^2-b^2-\dfrac{49b^2}{25}=433$

$a^2-\dfrac{7b^2}{5}=472$

Subtract them:

$\dfrac{39b^2}{25}=39 \Longrightarrow b=5$

Obtain $c$ :

$c=\dfrac{7(5)}{5}=7$

And obtain $a$ :

$a^2=(5)(7)+472 \Longrightarrow a=13\sqrt{3}$

Hence, $a+b+c=12+13\sqrt{3}$ , $d=12$ , $f=13$ and $df=\boxed{156}$ .

at least a or b or c is $p*\sqrt{3}$ from following equation,

we can know $a=p \times \sqrt{3}$ and b,c are natural number ---- (I) or

$b=p \times \sqrt{3}$ , $c=q \times \sqrt{3}$ , and a is natural number.-----(II)

and $b \leq c$ . (they can not be a rational number.)

I) From third equation

$(c-b)^2 = 39 - cb$

$(c-b)^2 \in \{ 1, 4, 9, 25, 36 \}$

$(b,c) \in \{ (2,7) , (5,7) \}$ can satisfy third equation.

(2,7) -> Number 'a' can't be $p \times \sqrt{3}$ form.

(5,7) -> $a = 13\sqrt{3}$

II) p,q are integer number because of third equation. so

$(b,c) \in { (3\sqrt{3},4\sqrt{3}) , (\sqrt{3}, 4\sqrt{3}) }$ can satisfy third equation.

but these case can't make 'a' natural number.

so $a=13\sqrt{3} , b = 5, c=7$

$a+b+c = 12 + 13\sqrt{3} = d + f\sqrt{3}$

$d \times f = 156$