Answer
Yes, these vectors are linearly independent.
Work Step by Step
By Theorem 4, $\det A=0$ if and only if $A$ is not invertible, and by the Invertible Matrix Theorem, a square matrix is invertible if and only if its columns are linearly independent. Hence, we can form a square matrix from the given vectors and compute the determinant in order to decide their linear independence.
$\begin{vmatrix}7&-8&7\\-4&5&0\\-6&7&-5\end{vmatrix}=7\begin{vmatrix}5&0\\7&-5\end{vmatrix}-(-8)\begin{vmatrix}-4&0\\-6&-5\end{vmatrix}+7\begin{vmatrix}-4&5\\-6&7\end{vmatrix}=7\cdot (-25)+8\cdot20+7\cdot 2=-1$
Since the determinant is nonzero, the vectors that form the columns of the matrix must be linearly independent.
