Calculate the determinant of M:

To determine if the vectors $x_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$, $x_2 = \begin{bmatrix} 3 \\ 2 \\ 9 \end{bmatrix}$, and $x_3 = \begin{bmatrix} 5 \\ 2 \\ -1 \end{bmatrix}$ are linearly independent, we can form a matrix $M$ with these vectors as columns and calculate its determinant. If the determinant is non-zero, the vectors are linearly independent.

$$M = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 2 & 2 \\ 3 & 9 & -1 \end{bmatrix}$$

Calculate the determinant of $M$:

$$\det(M) = 1 \cdot \det \begin{bmatrix} 2 & 2 \\ 9 & -1 \end{bmatrix} - 3 \cdot \det \begin{bmatrix} 2 & 2 \\ 3 & -1 \end{bmatrix} + 5 \cdot \det \begin{bmatrix} 2 & 2 \\ 3 & 9 \end{bmatrix}$$ $$\det(M) = 1 \cdot ((2)(-1) - (2)(9)) - 3 \cdot ((2)(-1) - (2)(3)) + 5 \cdot ((2)(9) - (2)(3))$$ $$\det(M) = 1 \cdot (-2 - 18) - 3 \cdot (-2 - 6) + 5 \cdot (18 - 6)$$ $$\det(M) = 1 \cdot (-20) - 3 \cdot (-8) + 5 \cdot (12)$$ $$\det(M) = -20 + 24 + 60$$ $$\det(M) = 64$$

Since $\det(M) = 64 \neq 0$, the vectors are linearly independent.
The set of vectors is $\boxed{\text{linearly independent}}$.

To determine if the vectors $x_1 = \begin{bmatrix} 4 \\ -1 \\ 2 \end{bmatrix}$, $x_2 = \begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix}$, and $x_3 = \begin{bmatrix} 5 \\ 2 \\ 4 \end{bmatrix}$ are linearly independent, we form a matrix $M$ with these vectors as columns and calculate its determinant.

$$M = \begin{bmatrix} 4 & 3 & 5 \\ -1 & 2 & 2 \\ 2 & 3 & 4 \end{bmatrix}$$

Calculate the determinant of $M$:

$$\det(M) = 4 \cdot \det \begin{bmatrix} 2 & 2 \\ 3 & 4 \end{bmatrix} - 3 \cdot \det \begin{bmatrix} -1 & 2 \\ 2 & 4 \end{bmatrix} + 5 \cdot \det \begin{bmatrix} -1 & 2 \\ 2 & 3 \end{bmatrix}$$ $$\det(M) = 4 \cdot ((2)(4) - (2)(3)) - 3 \cdot ((-1)(4) - (2)(2)) + 5 \cdot ((-1)(3) - (2)(2))$$ $$\det(M) = 4 \cdot (8 - 6) - 3 \cdot (-4 - 4) + 5 \cdot (-3 - 4)$$ $$\det(M) = 4 \cdot (2) - 3 \cdot (-8) + 5 \cdot (-7)$$ $$\det(M) = 8 + 24 - 35$$ $$\det(M) = 32 - 35$$ $$\det(M) = -3$$

Since $\det(M) = -3 \neq 0$, the vectors are linearly independent.