Verify the Mean Value Theorem for the function $f(x) = x^{3} - 5x^{2} - 3x$ in the interval $[1, 3]$. Find all $c \in (1, 3)$ such that $f^{\prime}(c) = \frac{f(3) - f(1)}{3 - 1}$.

(C) The given function is $f(x) = x^{3} - 5x^{2} - 3x$.
Since $f(x)$ is a polynomial,it is continuous on $[1, 3]$ and differentiable on $(1, 3)$.
The derivative is $f^{\prime}(x) = 3x^{2} - 10x - 3$.
Calculate the values at the endpoints:
$f(1) = (1)^{3} - 5(1)^{2} - 3(1) = 1 - 5 - 3 = -7$.
$f(3) = (3)^{3} - 5(3)^{2} - 3(3) = 27 - 45 - 9 = -27$.
The slope of the secant line is $\frac{f(3) - f(1)}{3 - 1} = \frac{-27 - (-7)}{2} = \frac{-20}{2} = -10$.
According to the Mean Value Theorem,there exists $c \in (1, 3)$ such that $f^{\prime}(c) = -10$.
$3c^{2} - 10c - 3 = -10$.
$3c^{2} - 10c + 7 = 0$.
Factoring the quadratic: $3c^{2} - 3c - 7c + 7 = 0 \Rightarrow 3c(c - 1) - 7(c - 1) = 0$.
$(3c - 7)(c - 1) = 0$.
This gives $c = 1$ or $c = \frac{7}{3}$.
Since $c \in (1, 3)$,the only valid value is $c = \frac{7}{3}$.