To determine if points $A(-2, 3, 5)$, $B(1, 2, 3)$, and $C(7, 0, -1)$ are collinear and to find the ratio in which point $C$ divides the line segment $AB$, we can follow these steps:

Step 1: Check for Collinearity

Points $A$, $B$, and $C$ are collinear if the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel, which happens if one vector is a scalar multiple of the other.

1. Find the vector $\overrightarrow{AB}$: $\overrightarrow{AB} = B - A = (1 - (-2), 2 - 3, 3 - 5) = (3, -1, -2)$

2. Find the vector $\overrightarrow{AC}$: $\overrightarrow{AC} = C - A = (7 - (-2), 0 - 3, -1 - 5) = (9, -3, -6)$

3. Check if $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel: If $\overrightarrow{AC} = k \cdot \overrightarrow{AB}$ for some scalar $k$, the vectors are parallel. Dividing component-wise: $k = \frac{9}{3} = \frac{-3}{-1} = \frac{-6}{-2} = 3$ Since all components give the same scalar $k = 3$, the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel, indicating that points $A$, $B$, and $C$ are collinear.

Step 2: Find the Ratio in Which $C$ Divides $AB$

Since $C$ is on the line segment $AB$, we can use the section formula to determine the ratio $k : 1$ in which $C$ divides $AB$.

If $C$ divides $AB$ in the ratio $k : 1$, then the coordinates of $C$ can be given by: $C = \left( \frac{k \cdot x_B + x_A}{k + 1}, \frac{k \cdot y_B + y_A}{k + 1}, \frac{k \cdot z_B + z_A}{k + 1} \right)$ Substitute $C(7, 0, -1)$, $A(-2, 3, 5)$, and $B(1, 2, 3)$ into the formula and solve for $k$.

1. For the $x$-coordinate: $7 = \frac{k \cdot 1 + (-2)}{k + 1}$ Multiply by $k + 1$: $7(k + 1) = k - 2$ $7k + 7 = k - 2$ $6k = -9 \Rightarrow k = 3$ Thus, $C$ divides $AB$