Let $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k},\ \vec{b} = 3\hat{i} + 2\hat{j} + 2\hat{k},\ \vec{c} = 4\hat{i} - 3\hat{j} + \hat{k}$, then the vectors $\vec{a},\ \vec{b},\ \vec{c}$ are

The vectors are $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$, $\vec{b} = 3\hat{i} + 2\hat{j} + 2\hat{k}$, and $\vec{c} = 4\hat{i} - 3\hat{j} + \hat{k}$.

To check for linear independence, we evaluate the scalar triple product $[\vec{a}\ \vec{b}\ \vec{c}]$ which is given by the determinant of their components:

$[\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 2 & -3 & 1 \\ 3 & 2 & 2 \\ 4 & -3 & 1 \end{vmatrix}$

Applying the row operation $R_3 \rightarrow R_3 - R_1$:

$[\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 2 & -3 & 1 \\ 3 & 2 & 2 \\ 2 & 0 & 0 \end{vmatrix}$

Expanding along the third row:

$[\vec{a}\ \vec{b}\ \vec{c}] = 2((-3)(2) - (1)(2)) = 2(-6 - 2) = -16$

Since $[\vec{a}\ \vec{b}\ \vec{c}] \neq 0$, the vectors are non-coplanar and hence linearly independent.

Answer: Linearly independent