Answer
The vectors are not linearly indepenent
Work Step by Step
Make $A$ be an $n \times n$ matrix. Theorem 4 says that $\operatorname{det}(A) \neq 0$ if and only if $f$ $A$ is invertible. The Invertible Matrix Theorem tells us that $A$ is invertible if and only if the columns of $A$ are linearly independent. So, we see that
$\operatorname{det}(A) \neq 0$ if and only if the columns of $A$ are linearly independent.
This then gives us a way to check if the given vectors are linearly independent:
we make a matrix with these vectors as columns and check if the determinant of that matrix equals 0 or not.
First, we'll expand along the fourth column:
\[
\left|\begin{array}{cccc}
3 & 2 & -2 & 0 \\
5 & -6 & -1 & 0 \\
-6 & 0 & 3 & 0 \\
4 & 7 & 0 & -2
\end{array}\right|=-2\left|\begin{array}{ccc}
3 & 2 & -2 \\
5 & -6 & -1 \\
-6 & 0 & 3
\end{array}\right|
\]
Then we'll row reduce, keeping track of how that changes the determinant:
\[
\begin{array}{l}
\begin{aligned}
\Delta &=-2\left|\begin{array}{ccc}
3 & 2 & -2 \\
5 & -6 & -1 \\
-6 & 0 & 3
\end{array}\right| \\
&=-2\left|\begin{array}{ccc}
3 & 2 & -2 \\
5 & -6 & -1 \\
-1 & -6 & 2
\end{array}\right|
\end{aligned} \\
= & \begin{array}{rl}
-1 & -6 & 2 \\
3 & 2 & -2 \\
5 & -6 & -1
\end{array} \mid \\
& =2 \begin{array}{ll}
-1 & -6 & 2 \\
0 & -16 & 4 \\
0 & -36 & 9
\end{array} \mid \\
& =2(-4)(-9)\left|\begin{array}{ccc}
-1 & -6 & 2 \\
0 & 4 & -1 \\
0 & 4 & -1
\end{array}\right| \\
& =2(-4)(-9)\left|\begin{array}{ccc}
-1 & -6 & 2 \\
0 & 4 & -1 \\
0 & 0 & 0
\end{array}\right| \\
& =2(-4)(-9)[(-1)(4)(0)]
\end{array}
\]
since the determinant is equal to $0,$ we find that the 4 vectors are linear dependent
