17 Given that the three interior angles $A, B, C$ of $\triangle ABC$ are opposite to the sides $a, b, c$ respectively, the vectors $\vec{m}=(1,1-\sqrt{3} \sin A), \vec{n}=(\cos A, 1)$, and $\vec{m} \perp \vec{n}$.
(1) Find the angle $A$;
(2) If $b+c=\sqrt{3} a$, find the value of $\sin \left(B+\frac{\pi}{6}\right)$.
(17) (1) Since $\vec{m} \perp \vec{n}$, we have $\vec{m} \cdot \vec{n}=0$, which means
$$
\cos A+1-\sqrt{3} \sin A=0,
$$

which simplifies to
$$
\sqrt{3} \sin A-\cos A=1,
$$

thus
$$
\sin \left(A-\frac{\pi}{6}\right)=\frac{1}{2},
$$

Since $0<A<\pi$, we have
$$
-\frac{\pi}{6}<A-\frac{\pi}{6}<\frac{5 \pi}{6},
$$

which implies $A-\frac{\pi}{6}=\frac{\pi}{6}$, so $A=\frac{\pi}{3}$.
(2) Since $b+c=\sqrt{3} a$, by the Law of Sines we get
$$
\sin B+\sin C=\sqrt{3} \sin A=\frac{3}{2} .
$$

Since $B+C=\frac{2 \pi}{3}$, we have
$$
\begin{array}{c}
\sin B+\sin \left(\frac{2 \pi}{3}-B\right)=\frac{3}{2}, \\
\frac{\sqrt{3}}{2} \cos B+\frac{3}{2} \sin B=\frac{3}{2},
\end{array}
$$

which simplifies to $\sin \left(B+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$.