Find $\int_{C} x^{2} z d s$, where $C$ is the line segment from $(0,1,-5)$ to $(-4,-1,-8)$.

Step 1 ：We are given the scalar field \(f(x, y, z) = x^{2} z\) and the curve \(C\) which is the line segment from \((0,1,-5)\) to \((-4,-1,-8)\).

Step 2 ：We can parametrize this line segment as follows: \(x(t) = 0 - 4t\), \(y(t) = 1 - 2t\), \(z(t) = -5 - 3t\), for \(0 \leq t \leq 1\).

Step 3 ：Substituting these into the formula for the line integral of a scalar field, we get \(\int_{0}^{1} f(x(t), y(t), z(t)) \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} dt\).

Step 4 ：Substituting the parametric equations into the scalar field, we get \(f = 16t^{2}(-3t - 5)\).

Step 5 ：Computing the derivative of each parametric equation, we get \(x' = -4\), \(y' = -2\), and \(z' = -3\).

Step 6 ：Substituting these into the formula for \(ds\), we get \(ds = \sqrt{29}\).

Step 7 ：Substituting these into the integral, we get \(\int_{0}^{1} -116t^{2}\sqrt{29} dt\).

Step 8 ：Computing the integral, we get \(-\frac{116\sqrt{29}}{3}\).

Step 9 ：Thus, the value of the line integral is \(\boxed{-\frac{116\sqrt{29}}{3}}\).