Answer: 156

Small Hint:

Use $B = (0,0),$ $C = (14,0),$ $A = (5,12);$ the circumcenter is $\left(7, \frac{33}{8}\right),$ and the rotation keeps all six vertices on the circumcircle

Big Hint:

Rotate direction $(3,-4)$ of $\overline{AC}$ to vertical; the side condition selects $\cos\varphi = \frac{4}{5},$ $\sin\varphi = -\frac{3}{5}.$ Use the shoelace formula

Solution:

Place $B = (0,0),$ $C = (14,0),$ $A = (5,12).$ The circumcenter lies on $x = 7,$ and equating distances to $B$ and $A$ gives $O = \left(7, \frac{33}{8}\right).$ The direction of $\overline{AC}$ is $C - A = (9, -12),$ parallel to $(3, -4).$ A rotation through $\varphi$ makes $\overline{A'C'}$ vertical exactly when it sends $(3,-4)$ to $(0, \pm 5),$ so $(\cos\varphi, \sin\varphi) = \left(\frac{4}{5}, -\frac{3}{5}\right)$ or $\left(-\frac{4}{5}, \frac{3}{5}\right).$ Use the scalar cross product to test sides of the directed line $B'C'.$ For the first rotation, $$\begin{aligned} (C'-B') \times (A'-B') &= 168, \\ (C'-B') \times (B-B') &= -\frac{189}{4}, \end{aligned}$$ while for the second rotation these quantities are $168$ and $\frac{651}{4},$ respectively. Thus $A'$ and $B$ are on opposite sides only for $\cos\varphi = \frac{4}{5},$ $\sin\varphi = -\frac{3}{5}.$

With this rotation, $P' = O + R(P - O)$ gives $$A' = \left(\tfrac{81}{8}, \tfrac{93}{8}\right),$$ $$B' = \left(-\tfrac{43}{40}, \tfrac{201}{40}\right),$$ $$C' = \left(\tfrac{81}{8}, -\tfrac{27}{8}\right).$$ For example, $A - O = \left(-2, \tfrac{63}{8}\right)$ rotates to $\left(\tfrac{25}{8}, \tfrac{15}{2}\right),$ giving $A' = \left(\tfrac{81}{8}, \tfrac{93}{8}\right).$

The hexagon $A A' C C' B B'$ is simple with these vertices in order, so the shoelace formula on $(5,12),$ $\left(\tfrac{81}{8}, \tfrac{93}{8}\right),$ $(14,0),$ $\left(\tfrac{81}{8}, -\tfrac{27}{8}\right),$ $(0,0),$ $\left(-\tfrac{43}{40}, \tfrac{201}{40}\right)$ gives area $\frac{1557}{10} = 155.7.$ The closest integer is $156.$