Triangle $ ABC$ obeys $ AB = 2AC$ and $ \angle{BAC} = 120^{\circ}.$ Points $ P$ and $ Q$ lie on segment $ BC$ such that
\begin{eqnarray*}
AB^2 + BC \cdot CP = BC^2 \\
3AC^2 + 2BC \cdot CQ = BC^2
\end{eqnarray*}
Find $ \angle{PAQ}$ in degrees.
1. **Given Conditions and Initial Setup:**
   - Triangle \( ABC \) with \( AB = 2AC \) and \( \angle BAC = 120^\circ \).
   - Points \( P \) and \( Q \) lie on segment \( BC \) such that:
     \[
     AB^2 + BC \cdot CP = BC^2
     \]
     \[
     3AC^2 + 2BC \cdot CQ = BC^2
     \]

2. **Assigning Variables:**
   - Let \( AB = 2c \) and \( AC = c \).
   - Using the Law of Cosines in \( \triangle ABC \):
     \[
     BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(120^\circ)
     \]
     \[
     BC^2 = (2c)^2 + c^2 - 2 \cdot 2c \cdot c \cdot (-\frac{1}{2})
     \]
     \[
     BC^2 = 4c^2 + c^2 + 2c^2 = 7c^2
     \]
     \[
     BC = \sqrt{7}c
     \]

3. **Finding \( CP \) and \( CQ \):**
   - From the given conditions:
     \[
     AB^2 + BC \cdot CP = BC^2
     \]
     \[
     (2c)^2 + \sqrt{7}c \cdot CP = 7c^2
     \]
     \[
     4c^2 + \sqrt{7}c \cdot CP = 7c^2
     \]
     \[
     \sqrt{7}c \cdot CP = 3c^2
     \]
     \[
     CP = \frac{3c}{\sqrt{7}}
     \]

   - Similarly for \( CQ \):
     \[
     3AC^2 + 2BC \cdot CQ = BC^2
     \]
     \[
     3c^2 + 2\sqrt{7}c \cdot CQ = 7c^2
     \]
     \[
     2\sqrt{7}c \cdot CQ = 4c^2
     \]
     \[
     CQ = \frac{2c}{\sqrt{7}}
     \]

4. **Calculating Distances \( PQ \) and \( PB \):**
   - Since \( CP > CQ \), \( Q \) lies between \( B \) and \( P \):
     \[
     PQ = CP - CQ = \frac{3c}{\sqrt{7}} - \frac{2c}{\sqrt{7}} = \frac{c}{\sqrt{7}}
     \]
     \[
     PB = BC - CP = \sqrt{7}c - \frac{3c}{\sqrt{7}} = \frac{4c}{\sqrt{7}}
     \]

5. **Using Stewart's Theorem:**
   - For \( AP \):
     \[
     c^2 \cdot CP + (2c)^2 \cdot PB = PA^2 \cdot BC + BC \cdot PB \cdot CP
     \]
     \[
     c^2 \cdot \frac{3c}{\sqrt{7}} + 4c^2 \cdot \frac{4c}{\sqrt{7}} = PA^2 \cdot \sqrt{7}c + \sqrt{7}c \cdot \frac{4c}{\sqrt{7}} \cdot \frac{3c}{\sqrt{7}}
     \]
     \[
     \frac{3c^3}{\sqrt{7}} + \frac{16c^3}{\sqrt{7}} = PA^2 \cdot \sqrt{7}c + \frac{12c^3}{7}
     \]
     \[
     \frac{19c^3}{\sqrt{7}} = PA^2 \cdot \sqrt{7}c + \frac{12c^3}{7}
     \]
     \[
     PA^2 = \frac{2c^2}{7}
     \]
     \[
     PA = \frac{2c}{\sqrt{7}}
     \]

   - For \( AQ \):
     \[
     c^2 \cdot CQ + (2c)^2 \cdot QB = QA^2 \cdot BC + BC \cdot QB \cdot CQ
     \]
     \[
     c^2 \cdot \frac{2c}{\sqrt{7}} + 4c^2 \cdot \frac{3c}{\sqrt{7}} = QA^2 \cdot \sqrt{7}c + \sqrt{7}c \cdot \frac{3c}{\sqrt{7}} \cdot \frac{2c}{\sqrt{7}}
     \]
     \[
     \frac{2c^3}{\sqrt{7}} + \frac{12c^3}{\sqrt{7}} = QA^2 \cdot \sqrt{7}c + \frac{6c^3}{7}
     \]
     \[
     \frac{14c^3}{\sqrt{7}} = QA^2 \cdot \sqrt{7}c + \frac{6c^3}{7}
     \]
     \[
     QA^2 = \frac{3c^2}{7}
     \]
     \[
     QA = \frac{\sqrt{3}c}{\sqrt{7}}
     \]

6. **Finding \( \angle PAQ \):**
   - Using the Law of Cosines in \( \triangle PAQ \):
     \[
     PQ^2 = PA^2 + AQ^2 - 2 \cdot PA \cdot AQ \cdot \cos(\angle PAQ)
     \]
     \[
     \left(\frac{c}{\sqrt{7}}\right)^2 = \left(\frac{2c}{\sqrt{7}}\right)^2 + \left(\frac{\sqrt{3}c}{\sqrt{7}}\right)^2 - 2 \cdot \frac{2c}{\sqrt{7}} \cdot \frac{\sqrt{3}c}{\sqrt{7}} \cdot \cos(\angle PAQ)
     \]
     \[
     \frac{c^2}{7} = \frac{4c^2}{7} + \frac{3c^2}{7} - \frac{4\sqrt{3}c^2}{7} \cdot \cos(\angle PAQ)
     \]
     \[
     \frac{c^2}{7} = \frac{7c^2}{7} - \frac{4\sqrt{3}c^2}{7} \cdot \cos(\angle PAQ)
     \]
     \[
     \frac{c^2}{7} = c^2 - \frac{4\sqrt{3}c^2}{7} \cdot \cos(\angle PAQ)
     \]
     \[
     \frac{1}{7} = 1 - \frac{4\sqrt{3}}{7} \cdot \cos(\angle PAQ)
     \]
     \[
     \frac{4\sqrt{3}}{7} \cdot \cos(\angle PAQ) = \frac{6}{7}
     \]
     \[
     \cos(\angle PAQ) = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}
     \]
     \[
     \angle PAQ = 30^\circ
     \]

The final answer is \(\boxed{30^\circ}\).