Calculate the determinant and inverse of the given 3x3 matrix A.

$$A = \begin{pmatrix} 2 & 1 & 2 \\ 3 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}$$

The determinant is calculated as:

$$\det(A) = 2 \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} + 2 \begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix}$$ $$\det(A) = 2((1)(1) - (2)(2)) - 1((3)(1) - (2)(2)) + 2((3)(2) - (1)(2))$$ $$\det(A) = 2(1 - 4) - 1(3 - 4) + 2(6 - 2)$$ $$\det(A) = 2(-3) - 1(-1) + 2(4)$$ $$\det(A) = -6 + 1 + 8$$ $$\boxed{\det(A) = 3}$$

The cofactor $C_{ij}$ is $(-1)^{i+j}$ times the determinant of the submatrix obtained by removing row $i$ and column $j$.

$$C_{11} = \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = 1 - 4 = -3$$ $$C_{12} = -\begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = -(3 - 4) = 1$$ $$C_{13} = \begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix} = 6 - 2 = 4$$ $$C_{21} = -\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = -(1 - 4) = 3$$ $$C_{22} = \begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix} = 2 - 4 = -2$$ $$C_{23} = -\begin{vmatrix} 2 & 1 \\ 2 & 2 \end{vmatrix} = -(4 - 2) = -2$$ $$C_{31} = \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 2 - 2 = 0$$ $$C_{32} = -\begin{vmatrix} 2 & 2 \\ 3 & 2 \end{vmatrix} = -(4 - 6) = 2$$ $$C_{33} = \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = 2 - 3 = -1$$

The cofactor matrix is:

$$C = \begin{pmatrix} -3 & 1 & 4 \\ 3 & -2 & -2 \\ 0 & 2 & -1 \end{pmatrix}$$

The adjugate matrix is the transpose of the cofactor matrix:

$$\text{adj}(A) = C^T = \begin{pmatrix} -3 & 3 & 0 \\ 1 & -2 & 2 \\ 4 & -2 & -1 \end{pmatrix}$$

The inverse matrix $A^{-1}$ is $\frac{1}{\det(A)} \text{adj}(A)$:

$$A^{-1} = \frac{1}{3} \begin{pmatrix} -3 & 3 & 0 \\ 1 & -2 & 2 \\ 4 & -2 & -1 \end{pmatrix}$$ $$\boxed{A^{-1} = \begin{pmatrix} -1 & 1 & 0 \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{4}{3} & -\frac{2}{3} & -\frac{1}{3} \end{pmatrix}}$$

Solve the system of equations

The given system is:

$$2x + y + 2z = 3$$ $$3x + y + 2z = 3$$ $$2x + 2y + z = 2$$

This can be written as:

$$\begin{pmatrix} 2 & 1 & 2 \\ 3 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}$$

Here, $A = \begin{pmatrix} 2 & 1 & 2 \\ 3 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, and $B = \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}$.

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{3} \begin{pmatrix} -3 & 3 & 0 \\ 1 & -2 & 2 \\ 4 & -2 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}$$

Perform the matrix multiplication:

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{3} \begin{pmatrix} (-3)(3) + (3)(3) + (0)(2) \\ (1)(3) + (-2)(3) + (2)(2) \\ (4)(3) + (-2)(3) + (-1)(2) \end{pmatrix}$$ $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{3} \begin{pmatrix} -9 + 9 + 0 \\ 3 - 6 + 4 \\ 12 - 6 - 2 \end{pmatrix}$$ $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{3} \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}$$ $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{1}{3} \\ \frac{4}{3} \end{pmatrix}$$

The solution to the system of equations is:

$$\boxed{x = 0, y = \frac{1}{3}, z = \frac{4}{3}}$$