Let $ O,H,N$ be the circumcenter, orthocenter and 9-point center of $ \triangle ABC.$ Assume that $ OH$ is parallel to the internal bisector of $ \angle BAC.$ Which implies that $ HO$ bisects $ \angle BHC,$ since $ HB,HC$ and the external bisector of $ \angle BAC$ bound an isosceles triangle with apex $H.$ Hence if $ U,V$ denote the midpoints of $ HC,HB,$ then the quadrilateral $ HUNV$ is cyclic, where $ N$ is the midpoint of the arc $ UV$ of its circumcircle. By simple angle chase we obtain then
$ \angle UNV + \angle BHC = 180^{\circ} \Longrightarrow 2\angle BHC + \angle BHC = 180^{\circ}$
$ \Longrightarrow \angle BHC = 60^{\circ} \Longrightarrow \angle BAC = 120^{\circ}.$