To solve this system of equations using the given inverse of matrix $A$, the solution can be calculated as:

$\vec{x} = A^{-1} \cdot \vec{b}$

Step-by-step solution:

1. Matrix Multiplication:

   \begin{bmatrix} 5 & 7 & -6 \\ 11 & 16 & -14 \\ 7 & 10 & -9 \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 0 \\ -7 \\ -8 \end{bmatrix}.$$ Compute the product $$A^{-1} \cdot \vec{b}$$: $$A^{-1} \cdot \vec{b} = \begin{bmatrix} 5 & 7 & -6 \\ 11 & 16 & -14 \\ 7 & 10 & -9 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ -7 \\ -8 \end{bmatrix}.$$

2. Perform Element-wise Calculation: $\text{Row 1: } 5(0) + 7(-7) + (-6)(-8) = -49 + 48 = -1.$ $\text{Row 2: } 11(0) + 16(-7) + (-14)(-8) = -112 + 112 = 0.$ $\text{Row 3: } 7(0) + 10(-7) + (-9)(-8) = -70 + 72 = 2.$

3. Solution Vector:

   \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}.$$

Final Answer:

$x = -1, \quad y = 0, \quad z = 2.$

Let me know if you want additional details on matrix multiplication or have other questions. Here are related questions to explore:

1. How is the inverse of a matrix calculated?
2. What conditions make a matrix invertible?
3. Can this problem be solved using other methods, such as Gaussian elimination?
4. What happens if the determinant of $A$ is zero?
5. How are systems of equations applied in real-world problems?

Tip: Always check the determinant of a matrix before attempting to find its inverse to ensure it is invertible.