$O$ and $I$ are the circumcentre and incentre of  $\vartriangle ABC$ respectively. Suppose $O$ lies in the interior of  $\vartriangle ABC$ and $I$ lies on the circle passing through $B, O$, and $C$. What is the magnitude of $\angle B AC$ in degrees?
1. **Identify the given information and the goal:**
   - \( O \) is the circumcenter of \( \triangle ABC \).
   - \( I \) is the incenter of \( \triangle ABC \).
   - \( O \) lies inside \( \triangle ABC \).
   - \( I \) lies on the circle passing through \( B, O, \) and \( C \).
   - We need to find the magnitude of \( \angle BAC \).

2. **Use the property of angles subtended by the same segment:**
   - Since \( I \) lies on the circle passing through \( B, O, \) and \( C \), the angles \( \angle BOC \) and \( \angle BIC \) are equal because they subtend the same arc \( BC \).

3. **Express \( \angle BOC \) in terms of \( \angle A \):**
   - \( O \) is the circumcenter, so \( \angle BOC \) is the central angle subtended by arc \( BC \).
   - The central angle \( \angle BOC \) is twice the inscribed angle \( \angle BAC \), i.e., \( \angle BOC = 2\angle A \).

4. **Express \( \angle BIC \) in terms of \( \angle A \):**
   - \( I \) is the incenter, so \( \angle BIC \) is given by the formula for the incenter angle:
     \[
     \angle BIC = 90^\circ + \frac{\angle A}{2}
     \]

5. **Set up the equation using the equality of \( \angle BOC \) and \( \angle BIC \):**
   - Since \( \angle BOC = \angle BIC \), we have:
     \[
     2\angle A = 90^\circ + \frac{\angle A}{2}
     \]

6. **Solve for \( \angle A \):**
   \[
   2\angle A = 90^\circ + \frac{\angle A}{2}
   \]
   Multiply both sides by 2 to clear the fraction:
   \[
   4\angle A = 180^\circ + \angle A
   \]
   Subtract \( \angle A \) from both sides:
   \[
   3\angle A = 180^\circ
   \]
   Divide by 3:
   \[
   \angle A = 60^\circ
   \]

The final answer is \(\boxed{60^\circ}\)