Take three points $P$, $Q$, $R$ inside of the cube in such a way that the quadrilaterals $KLMP$, $MNIQ$ and $IJKR$ are parallelograms. Since $IJ$ and $ML$ are parallel, so are $PK$ and $RK$, and we see that the points $P$, $R$, $K$ lie on a same straight line. Therefore, we conclude that the hexagon $IJKLMN$ is partitioned into 3 parallelograms and a triangle $PQR$ as indicated in the figure below.

We then have
$$\begin{aligned} \Delta IKM &= \Delta IKR + \Delta KMP + \Delta MIQ + \Delta PQR \\ &= \frac{1}{2}(\square IJKR + \square KLMP + \square MNIQ) + \Delta PQR \\ &= \frac{1}{2}(a - \Delta PQR) + \Delta PQR = \frac{1}{2}(a + \Delta PQR). \end{aligned}$$

Here we are denoting $\Delta XYZ$ and $\square XYZW$ for the areas of the triangle $XYZ$ and of the quadrilateral $XYZW$, respectively. We note also that since $P$ and $R$ lie on the same side from $K$ on the line $PK$, $PR = |PK - RK| = |ML - IJ|$ holds. Similarly, we have $PQ = |NI - LK|$ and $QR = |JK - MN|$.

Next, take three points $P'$, $Q'$, $R'$ inside of the cube so that the quadrilaterals $NIJP'$, $JKLQ'$, $LMNR'$ are parallelograms. Then, in the same way as above, we obtain that $\Delta JLN = \frac{1}{2}(a + \Delta P'Q'R')$. Furthermore, we get $P'R' = |ML - IJ| = PR$, $P'Q' = |NI - LK| = PQ$ and $Q'R' = |JK - MN| = QR$. Hence, the triangles $PQR$ and $P'Q'R'$ are congruent, and in particular, we have $\Delta PQR = \Delta P'Q'R'$. Thus we obtain the fact that the answer we seek is $\Delta IKM + \Delta JLN - a = \Delta PQR$, and thus it is enough to find the area of the triangle $PQR$.

Let us now observe that the right triangles $EJI$ and $CML$ are similar since the corresponding sides are parallel. Therefore, we have
$$\begin{aligned} EJ : EI : IJ &= (CM - EJ) : (CL - EI) : (LM - IJ) \\ &= 6 : ((2012 - GL) - (2012 - AI)) : (ML - IJ) = 6 : 8 : (ML - IJ). \end{aligned}$$
Since the triangle for which the ratios of the lengths of the three sides are $6 : 8 : |ML - IJ|$ is a right triangle whose hypotenuse is the side with length $|ML - IJ|$, we get $|ML - IJ| = \sqrt{6^2 + 8^2} = 10$. Thus, we get $PR = 10$. Similarly, we obtain $PQ = |NI - LK| = \sqrt{4^2 + 8^2} = 4\sqrt{5}$ and $QR = |JK - MN| = \sqrt{4^2 + 6^2} = 2\sqrt{13}$.

Let $S$ be the foot of the perpendicular line drawn from $Q$ to $PR$ and let $x = RS$ (assume that if $S$ lies on the opposite side from $P$ with respect to $R$, then $x$ is negative). Then, from the Pythagorean theorem, we get
$$(2\sqrt{13})^2 - x^2 = QS^2 = (4\sqrt{5})^2 - (10 - x)^2.$$
Solving for $x$ we obtain $x = \frac{18}{5}$, and then we get $QS = \frac{4\sqrt{61}}{5}$. Finally, we obtain
$$\Delta PQR = \frac{1}{2} \cdot 10 \cdot \frac{4\sqrt{61}}{5} = 4\sqrt{61},$$
which is the desired answer for the problem.