1. Identify the key points and properties:
- $\triangle ABC$ is an acute triangle with circumcenter $O$.
- $Q \neq A$ is a point on $\odot(ABC)$ such that $AQ \perp BC$.
- The circumcircle of $\triangle BOC$ intersects $AC$ and $AB$ at $D$ and $E$ respectively.
- $AQ$, $BC$, and $DE$ are concurrent.
- Given $OD = 3$ and $OE = 7$.

2. **Establish the cyclic nature of quadrilateral $ADQE$:**
- Since $Q$ lies on $\odot(ABC)$ and is collinear with $BE \cap CD$ and $BC \cap DE$, $Q$ is the Miquel point of the complete quadrilateral $\{BE, CD, BC, DE\}$.
- Therefore, quadrilateral $ADQE$ is cyclic.

3. **Determine the relationship between $O_2$ and $Q$:**
- Let $O_2$ be the circumcenter of $\odot(BOC)$.
- Since $O_2Q \perp AQ$, the antipode of $A$ with respect to $\odot(ABC)$, denoted as $K$, lies on line $QO_2$.

4. Analyze the orthocenter and circumradius properties:
- Given that $BC$ and $DE$ are antiparallel with respect to $\angle A$, $AO \perp DE$.
- Angle chasing shows that $O$ is the orthocenter of $\triangle ADE$.
- Thus, $\odot(ADE)$ and $\odot(BOC)$ have the same circumradius.

5. Set up the circumradius relationships:
- Let $R_1$ be the circumradius of $\odot(ABC)$ and $R_2$ be the circumradius of $\odot(ADE)$.
- We have $DO = 2R_2 \cos B = 3$ and $EO = 2R_2 \cos C = 7$.

6. **Calculate $AQ$:**
- $AQ = 2OO_2 = 2R_2$.
- Also, $AQ = AK \cos(B - C) = 2R_1 \cos(B - C)$.
- Using the relationship $R_2 = \frac{BC}{2 \sin 2A} = \frac{R_1}{2 \cos A}$, we get $\cos A \cos(B - C) = \frac{1}{2}$.

7. Perform trigonometric manipulations:
- $\cos A \cos(B - C) = \frac{1}{2}$.
- This implies $\cos(B + C) \cos(B - C) = -\frac{1}{2}$.
- Therefore, $\cos 2B + \cos 2C = -1$.

8. **Solve for $R_2$:**
- Using the given values, $2 \left(\frac{3}{2R_2}\right)^2 - 1 + 2 \left(\frac{7}{2R_2}\right)^2 - 1 = -1$.
- Simplifying, we get:
$$2 \left(\frac{9}{4R_2^2}\right) - 1 + 2 \left(\frac{49}{4R_2^2}\right) - 1 = -1$$
$$\frac{18}{4R_2^2} + \frac{98}{4R_2^2} - 2 = -1$$
$$\frac{116}{4R_2^2} - 2 = -1$$
$$\frac{29}{R_2^2} = 1$$
$$R_2^2 = 29$$
$$R_2 = \sqrt{29}$$

9. **Compute $AQ$:**
- $AQ = 2R_2 = 2\sqrt{29}$.

The final answer is $\boxed{2\sqrt{29}}$.