1. **Problem statement:** Determine $X$ and $Z$ such that points $A(-1,0,3)$, $B(3,-2,6)$, and $C(X,2,Z)$ are collinear. 2. **Formula and rule:** Three points are collinear if the vector $\overrightarrow{AB}$ is parallel to $\overrightarrow{AC}$. This means there exists a scalar $t$ such that: $$\overrightarrow{AC} = t \overrightarrow{AB}$$ 3. **Calculate vectors:** $$\overrightarrow{AB} = B - A = (3 - (-1), -2 - 0, 6 - 3) = (4, -2, 3)$$ $$\overrightarrow{AC} = C - A = (X - (-1), 2 - 0, Z - 3) = (X + 1, 2, Z - 3)$$ 4. **Set up proportionality:** Since $\overrightarrow{AC} = t \overrightarrow{AB}$, we have: $$X + 1 = 4t$$ $$2 = -2t$$ $$Z - 3 = 3t$$ 5. **Solve for $t$ from the second equation:** $$2 = -2t \implies t = -1$$ 6. **Find $X$ and $Z$ using $t = -1$:** $$X + 1 = 4(-1) = -4 \implies X = -5$$ $$Z - 3 = 3(-1) = -3 \implies Z = 0$$ 7. **Answer:** The points are collinear if $X = -5$ and $Z = 0$. **Final answer:** $X = -5$, $Z = 0$