In a triangle $ABC$, $P$ and $Q$ are the feet of the altitudes from $B$ and $A$ respectively. Find the locus of the circumcentre of triangle $PQC$, when point $C$ varies (with $A$ and $B$ fixed) in such a way that $\angle ACB$ is equal to $60^{\circ}$.
1. **Identify the orthocenter \( H \) of \(\triangle ABC\)**:
   - The orthocenter \( H \) is the intersection of the altitudes from vertices \( A \), \( B \), and \( C \). Specifically, \( H \) is the intersection of \( BP \) and \( AQ \).

2. **Establish that \( CPHQ \) is a cyclic quadrilateral**:
   - Since \( P \) and \( Q \) are the feet of the altitudes from \( B \) and \( A \) respectively, we have \(\angle CPH = \angle CQH = 90^\circ\).
   - Given \(\angle ACB = 60^\circ\), we can deduce that \(\angle PHQ = 180^\circ - \angle ACB = 120^\circ\).
   - Therefore, \( CPHQ \) is a cyclic quadrilateral because opposite angles sum to \(180^\circ\).

3. **Determine the locus of \( H \)**:
   - For all points \( C \) such that \(\angle ACB = 60^\circ\), \( C \) lies on the circumcircle \( O \) of \(\triangle ABC\).
   - Since \(\angle AHB = 2 \angle ACB = 120^\circ\), the locus of \( H \) is the reflection of the circumcircle \( O \) across the line \(\overline{AB}\). We denote this reflected circle as \( O' \).

4. **Analyze the relationship between \( CH \) and the circumcircle of \(\triangle CPQ\)**:
   - Since \(\overline{CH} \perp \overline{AB}\), the length \( CH \) is constant and equal to the distance between the centers of \( O \) and \( O' \).
   - The line segment \( \overline{CH} \) is the diameter of the circumcircle of \(\triangle CPQ\) because \(\angle CPH = 90^\circ\).

5. **Find the locus of the circumcenter \( R \) of \(\triangle CPQ\)**:
   - The circumcenter \( R \) of \(\triangle CPQ\) is the midpoint of \( \overline{CH} \).
   - Since \( CH \) is constant and equal to the distance between the centers of \( O \) and \( O' \), the locus of \( R \) is a circle with radius equal to half the distance between \( O \) and \( O' \).
   - The center of this circle is the midpoint of \( \overline{AB} \).

The final answer is the major arc of a circle whose radius is equal to the radius of the circumcircle of \( \boxed{ \triangle ABC } \), and with center located at the midpoint of \(\overline{AB}\).