Let $ABC$ be a triangle with $AB=5,AC=4,BC=6$. The internal angle bisector of $C$ intersects the side $AB$ at $D$. Points $M$ and $N$ are taken on sides $BC$ and $AC$, respectively, such that $DM\parallel AC$ and $DN\parallel BC$. If $(MN{)}^2=p/q$ where $p$ and $q$ are relatively prime positive integers, then what is the sum of the digits of $|p-q|$?

Step 1: By the Angle Bisector Theorem, $CD$ divides $AB$ in the ratio of the adjacent sides:

Since $AB=5$, we have $AD=2$ and $BD=3$.

Step 2: In the quadrilateral $CMDN$, $DM\parallel AC$ and $DN\parallel BC$, so $CMDN$ is a parallelogram. Using similar triangles $△BMD\sim △BCA$ and $△AND\sim △ACB$:

Since $DM=DN=12/5$, the parallelogram $CMDN$ is actually a rhombus, and $CN=CM=12/5$.

Step 3: Use the Law of Cosines on $△ABC$ to find $\cos C$:

Step 4: In $△CMN$, use the Law of Cosines to find $(MN{)}^2$:

Here, $p=126$ and $q=25$, which are relatively prime.

Step 5: Calculate $|p-q|$ and the sum of its digits:

The sum of the digits of $101$ is $1+0+1=2$.