Let $P$ be a point inside the equilateral triangle $ABC$ such that $6\angle  PBC = 3\angle  PAC = 2\angle  PCA$. Find the measure of the angle $\angle PBC$ .
1. Let \( x = \angle PBC \). Given the conditions \( 6\angle PBC = 3\angle PAC = 2\angle PCA \), we can express the angles in terms of \( x \):
   \[
   \angle PBC = x, \quad \angle PAC = 3x, \quad \angle PCA = 2x
   \]
   Since \( P \) is inside the equilateral triangle \( ABC \), the sum of the angles around point \( P \) must be \( 360^\circ \). Therefore:
   \[
   \angle PBC + \angle PAC + \angle PCA = 60^\circ
   \]
   Substituting the expressions for the angles, we get:
   \[
   x + 3x + 2x = 60^\circ \implies 6x = 60^\circ \implies x = 10^\circ
   \]

2. To verify this, we use the trigonometric form of Ceva's theorem. According to Ceva's theorem in trigonometric form, for point \( P \) inside triangle \( ABC \):
   \[
   \frac{\sin(\angle PBC)}{\sin(\angle PCA)} \cdot \frac{\sin(\angle PAC)}{\sin(\angle PAB)} \cdot \frac{\sin(\angle PBA)}{\sin(\angle PBC)} = 1
   \]
   Substituting the angles:
   \[
   \frac{\sin(x)}{\sin(60^\circ - 3x)} \cdot \frac{\sin(3x)}{\sin(60^\circ - x)} \cdot \frac{\sin(60^\circ - 2x)}{\sin(2x)} = 1
   \]

3. Simplify the equation using the identity \( \sin(60^\circ - \theta) = \sin(60^\circ - \theta) \):
   \[
   \frac{\sin(x)}{\sin(60^\circ - 3x)} \cdot \frac{\sin(3x)}{\sin(60^\circ - x)} \cdot \frac{\sin(60^\circ - 2x)}{\sin(2x)} = 1
   \]

4. Set \( t = \cos(2x) \). Using the identity \( \sin(2x) = 2\sin(x)\cos(x) \) and \( \cos(2x) = 1 - 2\sin^2(x) \), we transform the equation:
   \[
   t^2 + 3t - 4t^3 = \sqrt{3(1 - t^2)} \cdot t
   \]

5. Simplify further:
   \[
   (1 - t)(3 + 4t) = \sqrt{3(1 - t^2)}
   \]

6. Square both sides to eliminate the square root:
   \[
   (1 - t)^2(3 + 4t)^2 = 3(1 - t^2)
   \]

7. Solve for \( t \):
   \[
   t = -\frac{1}{2} \quad \text{or} \quad t = -\frac{\sqrt{3}}{2} \quad \text{or} \quad t = \frac{\sqrt{3}}{2}
   \]

8. Since \( t = \cos(2x) \) and \( t \in (0, 1) \), the valid solution is:
   \[
   \cos(2x) = \frac{\sqrt{3}}{2} \implies 2x = 30^\circ \implies x = 15^\circ
   \]

The final answer is \( \boxed{15^\circ} \)