Let \(\overset{⃗}{a} = 2\hat{i}-3\hat{j}+\hat{k}, \overset{⃗}{b} = 3\hat{i}+2\hat{j}+2\hat{k}, \overset{⃗}{c} = 4\hat{i}-3\hat{j}+\hat{k}\), then the vectors \(\overset{⃗}{a}, \overset{⃗}{b}, \overset{⃗}{c}\) are
Step 1: Understanding the Concept
Three vectors are coplanar (linearly dependent) exactly when their scalar triple product is zero.
Step 2: Key Formula or Approach
\[ [\vec a\ \vec b\ \vec c]=\begin{vmatrix}2&-3&1\\3&2&2\\4&-3&1\end{vmatrix} \]
Step 3: Detailed Explanation
Expand along the first row: \(2(2-(-6))-(-3)(3-8)+1(-9-8)=16-15-17=-16\).
The determinant is nonzero, so the vectors are linearly independent and not coplanar.
Also \(\vec a\cdot\vec b=6-6+2=2\neq0\), so \(\vec a,\vec b\) are not orthogonal.
Final Answer:
The vectors are linearly independent, option (C).
\[ \boxed{\text{Linearly independent (C)}} \]