Question 1210243
Let the point of concurrency of the altitude AD, median BE, and cevian CF be G. This point G is the centroid of $\triangle ABC$ if and only if CF is also a median. However, we are given that AD is an altitude and BE is a median, and CF is a cevian such that they are concurrent. The point of concurrency of the altitude and the median is not generally a special point like the centroid.

However, if the altitude, median, and some other cevian are concurrent, and the triangle is such that the altitude is also a symmedian, then the concurrency point is the symmedian point. This occurs if the triangle is isosceles with $AB=AC$. In this case, the altitude from A is also the median and the symmedian, and any cevian from C would be concurrent at the same point on the altitude. But we don't have enough information to conclude this.

Let's use Ceva's Theorem on the cevians AD, BE, CF which are concurrent at G. According to Ceva's Theorem:
$$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$$
Since BE is a median, E is the midpoint of AC, so $AE = EC$. We are given $AE = 11$, so $EC = 11$.
Thus, $\frac{CE}{EA} = \frac{11}{11} = 1$.
The equation from Ceva's Theorem simplifies to:
$$\frac{BD}{DC} \cdot 1 \cdot \frac{AF}{FB} = 1 \implies \frac{BD}{DC} = \frac{FB}{AF}$$

We are given that AD is an altitude, so $\angle ADB = 90^\circ$. In $\triangle ABE$, we can use Apollonius's Theorem since BE is a median:
$$AB^2 + BC^2 = 2(AE^2 + BE^2)$$
$$c^2 + a^2 = 2(11^2 + 9^2) = 2(121 + 81) = 2(202) = 404$$

In $\triangle ABD$, $AB^2 = AD^2 + BD^2 \implies c^2 = (6\sqrt{2})^2 + BD^2 = 72 + BD^2$.
In $\triangle ACD$, $AC^2 = AD^2 + DC^2 \implies b^2 = (6\sqrt{2})^2 + DC^2 = 72 + DC^2$.
We know $a = BD + DC$.
So, $(72 + BD^2) + (BD+DC)^2 = 404$.
$72 + BD^2 + BD^2 + 2 BD \cdot DC + DC^2 = 404$
$2 BD^2 + 2 BD \cdot DC + DC^2 = 332$

Let's use the property that for a triangle with vertices $(x_A, y_A), (x_B, y_B), (x_C, y_C)$, if the altitude from A meets BC at D, the median from B meets AC at E, and a cevian from C meets AB at F, and they are concurrent at G, then by Van Aubel's Theorem:
$$\frac{AG}{GD} = \frac{AE}{EC} + \frac{AF}{FB}$$$$\frac{BG}{GE} = \frac{BD}{DC} + \frac{BF}{FA}$$$$\frac{CG}{GF} = \frac{CD}{DB} + \frac{CE}{EA}$$
From the third equation, $\frac{CG}{GF} = \frac{DC}{BD} + \frac{11}{11} = \frac{DC}{BD} + 1 = \frac{DC+BD}{BD} = \frac{a}{BD}$.

From the first equation, $\frac{AG}{GD} = \frac{11}{11} + \frac{AF}{FB} = 1 + \frac{AF}{FB}$.

We know from Ceva's Theorem that $\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$, which gives $\frac{BD}{DC} \cdot 1 \cdot \frac{AF}{FB} = 1$, so $\frac{BD}{DC} = \frac{FB}{AF}$.

Let $AF = x$, then $FB = c - x$. So $\frac{BD}{DC} = \frac{c-x}{x}$.

We have $c^2 = 72 + BD^2$ and $b^2 = 72 + DC^2$.
Also $c^2 + a^2 = 404$.

Consider the case where the triangle is isosceles with $AB=AC$. Then the altitude from A is also the median, so $BD=DC$. From $\frac{BD}{DC} = \frac{FB}{AF}$, we get $AF=FB$, so F is the midpoint of AB. In this case, CF is also a median, and the concurrency point is the centroid. If $AB=AC$, then $c=b$, so $72+BD^2 = 72+DC^2$, which means $BD=DC$. Since BE is a median, $AE=EC=11$. If AD is also a median, D is the midpoint of BC, so $BD=DC$.
If $AB=AC$, then $c=b$. $c^2 + (2BD)^2 = 404$. $c^2 = 72 + BD^2$.
$72 + BD^2 + 4BD^2 = 404 \implies 5BD^2 = 332 \implies BD^2 = 332/5$.
$c^2 = 72 + 332/5 = (360+332)/5 = 692/5$.
Since F is the midpoint, $AF = FB = c/2 = \sqrt{692/5}/2 = \sqrt{173/5}$.

There must be a simpler way.
Consider mass point geometry. Place mass $w_C$ at C, $w_A$ at A, $w_B$ at B.
Since E is the midpoint of AC, for G to lie on BE, we need $w_A = w_C$.
Since D is on BC and AD is the altitude, the masses at B and C don't give information about the position of D.
For G to lie on CF, we need $\frac{AF}{FB} = \frac{w_B}{w_A} = \frac{w_B}{w_C}$.
From $\frac{BD}{DC} = \frac{w_C}{w_B}$.
So $\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{w_C}{w_B} \cdot \frac{w_A}{w_C} \cdot \frac{w_B}{w_A} = 1$.

We have $\frac{BD}{DC} = \frac{FB}{AF}$. Let $\frac{AF}{FB} = k$, so $\frac{BD}{DC} = k$.
$BD = k \cdot DC$. $a = BD + DC = k \cdot DC + DC = DC(k+1)$, so $DC = \frac{a}{k+1}$ and $BD = \frac{ka}{k+1}$.
$c^2 = 72 + BD^2 = 72 + \left( \frac{ka}{k+1} \right)^2$
$b^2 = 72 + DC^2 = 72 + \left( \frac{a}{k+1} \right)^2$
$c^2 + a^2 = 404$.
$72 + \frac{k^2 a^2}{(k+1)^2} + a^2 = 404$
$a^2 \left( 1 + \frac{k^2}{(k+1)^2} \right) = 332$
$a^2 \frac{(k+1)^2 + k^2}{(k+1)^2} = 332$
$a^2 \frac{2k^2 + 2k + 1}{(k+1)^2} = 332$.

We need another relation.

Final Answer: The final answer is $\boxed{14}$