Answer: 336

Small Hint:

The circle through the midpoints is the nine-point circle, so $G,$ $H,$ $J$ are the feet of the altitudes, and $\triangle DEF$ has the same angles as $\triangle ABC$

Big Hint:

Since $\angle BJC = \angle BHC = 90^\circ,$ $D$ is equidistant from $B,$ $C,$ $H,$ $J;$ isosceles triangles turn the arcs into angles of $\triangle ABC$

Solution:

The medial triangle $DEF$ has sides parallel to those of $ABC,$ so $\angle FDE = 84^\circ,$ $\angle DEF = 60^\circ,$ and $\angle DFE = 36^\circ.$ Its circumcircle is the nine-point circle, whose second intersections with the sides of $ABC$ are the feet of the altitudes: $G$ is the foot from $A,$ $H$ the foot from $B,$ and $J$ the foot from $C.$ By the inscribed angle theorem, $\overset{\frown}{DE} = 2\angle DFE = 72^\circ.$

For $\overset{\frown}{FG}:$ since $\overline{DF} \parallel \overline{CA}$ and $G$ lies on ray $DB,$ the angle $\angle FDG$ equals the angle between lines $CA$ and $CB,$ which is $36^\circ,$ so $\overset{\frown}{FG} = 2 \cdot 36^\circ = 72^\circ.$ For $\overset{\frown}{HJ}:$ because $\angle BJC = \angle BHC = 90^\circ,$ both $H$ and $J$ lie on the circle with diameter $\overline{BC}$ centered at $D,$ so $DJ = DB$ and $DH = DC.$ Isosceles triangle $BDJ$ gives $\angle JDB = 180^\circ - 2 \cdot 60^\circ = 60^\circ,$ and isosceles triangle $CDH$ gives $\angle HDC = 180^\circ - 2 \cdot 36^\circ = 108^\circ.$ Hence $\angle JDH = 180^\circ - 60^\circ$ $- 108^\circ = 12^\circ$ and $\overset{\frown}{HJ} = 24^\circ.$

Therefore $\overset{\frown}{DE} + 2 \cdot \overset{\frown}{HJ}$ $+ 3 \cdot \overset{\frown}{FG}$ $= 72 + 48 + 216 = 336.$