Joining

First we find the area of $\triangle ABC$ using Heron's formula, and it's: $[ABC]=6\sqrt{6}$ . Now, we find the length of the three excircles with the known formulas:

$R_a=\dfrac{2[ABC]}{-BC+AB+AC}=\dfrac{12\sqrt{6}}{4}=3\sqrt{6}$ $R_b=\dfrac{2[ABC]}{BC+AB-AC}=\dfrac{12\sqrt{6}}{6}=2\sqrt{6}$ $R_c=\dfrac{2[ABC]}{BC-AB+AC}=\dfrac{12\sqrt{6}}{8}=\dfrac{3\sqrt{6}}{2}$

With the lengths of the exradius, we find the area of $\triangle BDC$ , $\triangle CEA$ and $\triangle ABF$ , because the exradius are the altitudes and each side is the base:

$[BDC]=\dfrac{BC \times R_a}{2}=\dfrac{7 \times 3\sqrt{6}}{2}=\dfrac{21\sqrt{6}}{2}$ $[CEA]=\dfrac{AC \times R_b}{2}=\dfrac{6 \times 2\sqrt{6}}{2}=6\sqrt{6}$ $[ABF]=\dfrac{AB \times R_c}{2}=\dfrac{5 \times \frac{3\sqrt{6}}{2}}{2}=\dfrac{15\sqrt{6}}{4}$

Clearly, if we sum the four areas we obtained, we obtain the area of $\triangle DEF$ :

$[DEF]=[ABC]+[BDC]+[CEA]+[ABF]$ $[DEF]=6\sqrt{6}+\dfrac{21\sqrt{6}}{2}+6\sqrt{6}+\dfrac{15\sqrt{6}}{4}=\dfrac{105\sqrt{6}}{4}$

Next, with a little angle chasing, we find out that:

$\angle D=\dfrac{\angle B + \angle C}{2}$ , $\angle E=\dfrac{\angle A + \angle C}{2}$ and $\angle F=\dfrac{\angle A + \angle B}{2}$

Our goal now is to find the sines of every angle, to do that we first obtain the sines of $\triangle ABC$ using law of cosines, then apply the angle sum and finally the formula for a half angle. We obtain:

$\sin \angle D=\dfrac{\sqrt{15}}{5}$ , $\sin \angle E=\dfrac{3\sqrt{105}}{35}$ and $\sin \angle F=\dfrac{\sqrt{42}}{7}$

At this point we have the area of $\triangle DEF$ and every of its three angles. With law of sines and its formulas for the area of triangle, we will find every side:

$[DEF]=\dfrac{DE \times DF \times \sin \angle D}{2}$

$[DEF]=\dfrac{DE \times EF \times \sin \angle E}{2}$

$[DEF]=\dfrac{DF \times EF \times \sin \angle F}{2}$

Substitute with the known values and simplify:

$\dfrac{105\sqrt{6}}{4}=\dfrac{DE \times DF \times \sqrt{15}}{10} \Rightarrow DE \times DF=\dfrac{105 \sqrt{10}}{2}$

$\dfrac{105\sqrt{6}}{4}=\dfrac{DE \times EF \times 3\sqrt{105}}{70} \Rightarrow DE \times EF=\dfrac{35 \sqrt{70}}{2}$

$\dfrac{105\sqrt{6}}{4}=\dfrac{DF \times EF \times \sqrt{42}}{14} \Rightarrow DF \times EF=\dfrac{105 \sqrt{7}}{2}$

To solve that system, multiply the three equations, simplify and let the three sides to be in the left hand side:

$(DE \times DF \times EF)^2=\dfrac{13505625}{4}$

Take square root to both sides:

$DE \times DF \times EF = \dfrac{3675}{2}$

Finally, divide every of the three equations with this one to obtain every side:

$EF=\dfrac{\frac{3675}{2}}{\frac{105 \sqrt{10}}{2}}=\dfrac{7\sqrt{10}}{2}$

$DF=\dfrac{\frac{3675}{2}}{\frac{35 \sqrt{70}}{2}}=\dfrac{3\sqrt{70}}{2}$

$DE=\dfrac{\frac{3675}{2}}{\frac{105 \sqrt{7}}{2}}=5\sqrt{7}$

Hence, $a=10$ , $b=7$ , $c=7$ , $d=3$ , $f=5$ , $g=2$ and $abc-dfg=\boxed{460}$ .

Note: if you try to generalize it for sides $BC=a$ , $AC=b$ and $AB=c$ , you would obtain this interesting simple formulas:

$d=EF=\dfrac{2a\sqrt{bc}}{\sqrt{(a+b-c)(a-b+c)}}$ $e=DF=\dfrac{2b\sqrt{ac}}{\sqrt{(-a+b+c)(a+b-c)}}$ $f=DE=\dfrac{2c\sqrt{ab}}{\sqrt{(-a+b+c)(a-b+c)}}$

An important trick is to use here half-angle fomuales.

First note that $EF=EA+FA,DF=DB+FB,DE=DC+EC$

We drop the perpendiculars from $D,E,F$ on $BC,AC,AB$ respectively and name them $G,H,I$ respectively.

It is easy to see that $\angle HEA=\frac{A}{2} and \angle IFA=\frac{A}{2}$ and similarly for others.

So we have $EF=EGsec(\frac{A}{2})+FIsec(\frac{A}{2})=({r}_{2}+{r}_{3})sec(\frac{A}{2})$ (i)

We will use ${ r }_{ 1 }=s(tan\frac { A }{ 2 } ),{ r }_{ 2 }=s(tan\frac { B }{ 2 } ),{ r }_{ 3 }=s(tan\frac { C }{ 2 } )$ in (i) to get :

EF= $s(sec(\frac{A}{2}))((tan\frac { B }{ 2 } )+(tan\frac { C }{ 2 } ))$

We will use $tan\frac { B }{ 2 } =\frac { (s-a)(s-b) }{ \Delta } ,tan\frac { C }{ 2 } =\frac { (s-a)(s-b) }{ \Delta }$ to get

$EF=(sec\frac { A }{ 2 } )\frac { as(s-a) }{ \Delta } =asec\frac { A }{ 2 } cot\frac { A }{ 2 } =\frac { a }{ sin\frac { A }{ 2 } } \quad (Using\quad tan\frac { A }{ 2 } =\frac { \Delta }{ s(s-a) } )$

Similarly $DF=\frac { b }{ sin\frac { B }{ 2 } } ,DE=\frac { c }{ sin\frac { C }{ 2 } }$

Finally use $sin\frac { A }{ 2 } =\sqrt { \frac { (s-b)(s-c) }{ bc } } ,sin\frac { B }{ 2 } =\sqrt { \frac { (s-a)(s-c) }{ ac } } ,sin\frac { C }{ 2 } =\sqrt { \frac { (s-a)(s-b) }{ ab } }$ to get the values.