Consider the following equations:
$$
\begin{aligned}
3 \mathrm{~A}+6 \mathrm{~B} \longrightarrow & 3 \mathrm{D} & \Delta H &=-403 \mathrm{~kJ} / \mathrm{mol} \\
\mathrm{E}+2 \mathrm{~F} & \longrightarrow \mathrm{A} & \Delta H &=-105.2 \mathrm{~kJ} / \mathrm{mol} \\
\mathrm{C} & \longrightarrow \mathrm{E}+3 \mathrm{D} & \Delta H &=+64.8 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$
Suppose the first equation is reversed and multiplied by $\frac{1}{6}$, the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?