Let $H$ be the orthocenter of triangle $ABC$ and $K$ be the common orthocenter of triangle $ABQ$ and triangle $ACP$. Let $D$ be the foot of the perpendicular from $A$ to line $BC$. By definition, $A, D, H, K$ are collinear.

Since $AB = 10$, $AC = 11$ and $BC > 5$, both $\angle B$ and $\angle C$ are less than $90^\circ$. Therefore $D$ lies on side $BC$ and $D$ is different from $H$. Assume that $D$ and $K$ coincide, then two points $P$ and $Q$ coincide with $D$. The Pythagorean theorem shows that $AB^2 - BP^2 = AD^2 = AC^2 - CQ^2$, which is a contradiction because $10^2 - 5^2 \ne 11^2 - 6^2$. It follows that $D$ and $K$ are different points.

Two lines $BH$ and $PK$ are both perpendicular to side $AC$. Therefore these two lines are parallel and $DB : BP = DH : HK$ follows. Similarly, two lines $CH$ and $QK$ are parallel and $DC : CQ = DH : HK$ follows. From the above, it follows that $DB : BP = DC : CQ$ and $DB : DC = BP : CQ = 5 : 6$. Set $DB = 5t$, $DC = 6t$ where $t$ is a real number, then by using Pythagorean theorem for triangles $ABD$ and $ACD$ we have $10^2 - (5t)^2 = 11^2 - (6t)^2$. Since the required length is $BC = 11t$, the answer is $\sqrt{231}$.