Matrices $A$, $B$ are given as follows.
\[A=\begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 4 & 2 & 0 \\ 2 & 4 & 0 \\ 0 & 0 & 12\end{pmatrix}\]
Find volume of $V=\{\mathbf{x}\in\mathbb{R}^3 : \mathbf{x}\cdot A\mathbf{x} \leq 1 < \mathbf{x}\cdot B\mathbf{x} \}$.
To find the volume of the region \( V = \{\mathbf{x} \in \mathbb{R}^3 : \mathbf{x} \cdot A \mathbf{x} \leq 1 < \mathbf{x} \cdot B \mathbf{x} \} \), we need to understand the geometric interpretation of the given quadratic forms. The matrices \( A \) and \( B \) define ellipsoids in \(\mathbb{R}^3\).

1. **Simultaneous Orthogonal Diagonalization**:
   Since \( A \) and \( B \) are symmetric matrices, they can be orthogonally diagonalized. This means there exists an orthogonal matrix \( P \) such that:
   \[
   P^T A P = D_A \quad \text{and} \quad P^T B P = D_B
   \]
   where \( D_A \) and \( D_B \) are diagonal matrices containing the eigenvalues of \( A \) and \( B \), respectively.

2. **Finding Eigenvalues**:
   The eigenvalues of \( A \) and \( B \) are given as \(\{3, 3, 1\}\) and \(\{12, 6, 2\}\), respectively. This means:
   \[
   D_A = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad D_B = \begin{pmatrix} 12 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 2 \end{pmatrix}
   \]

3. **Volume of the Region Between Two Ellipsoids**:
   The volume of the region between two ellipsoids defined by \( \mathbf{x} \cdot A \mathbf{x} \leq 1 \) and \( \mathbf{x} \cdot B \mathbf{x} \leq 1 \) can be found by considering the volumes of the individual ellipsoids.

   The volume of an ellipsoid defined by \( \mathbf{x}^T D \mathbf{x} \leq 1 \) is given by:
   \[
   V = \frac{4\pi}{3} \frac{1}{\sqrt{\lambda_1 \lambda_2 \lambda_3}}
   \]
   where \(\lambda_1, \lambda_2, \lambda_3\) are the eigenvalues of the matrix \( D \).

4. **Calculating the Volumes**:
   For the ellipsoid defined by \( \mathbf{x} \cdot A \mathbf{x} \leq 1 \):
   \[
   V_A = \frac{4\pi}{3} \frac{1}{\sqrt{3 \cdot 3 \cdot 1}} = \frac{4\pi}{3} \frac{1}{\sqrt{9}} = \frac{4\pi}{3} \frac{1}{3} = \frac{4\pi}{9}
   \]

   For the ellipsoid defined by \( \mathbf{x} \cdot B \mathbf{x} \leq 1 \):
   \[
   V_B = \frac{4\pi}{3} \frac{1}{\sqrt{12 \cdot 6 \cdot 2}} = \frac{4\pi}{3} \frac{1}{\sqrt{144}} = \frac{4\pi}{3} \frac{1}{12} = \frac{\pi}{9}
   \]

5. **Volume of the Region Between the Ellipsoids**:
   The volume of the region \( V \) is the difference between the volumes of the two ellipsoids:
   \[
   V = V_A - V_B = \frac{4\pi}{9} - \frac{\pi}{9} = \frac{3\pi}{9} = \frac{\pi}{3}
   \]

The final answer is \(\boxed{\frac{\pi}{3}}\).