1. **State the problem:** Given points $A(-2,0,10)$, $B(1,9,3)$, and $C(2,-4,6)$, find vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$, then find the angle $\angle ACB$. 2. **Find vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$:** $$\overrightarrow{AC} = C - A = (2 - (-2), -4 - 0, 6 - 10) = (4, -4, -4)$$ $$\overrightarrow{BC} = C - B = (2 - 1, -4 - 9, 6 - 3) = (1, -13, 3)$$ 3. **Formula for angle between two vectors:** $$\cos \theta = \frac{\overrightarrow{AC} \cdot \overrightarrow{BC}}{|\overrightarrow{AC}| |\overrightarrow{BC}|}$$ where $\cdot$ is the dot product and $|\cdot|$ is the magnitude. 4. **Calculate dot product $\overrightarrow{AC} \cdot \overrightarrow{BC}$:** $$\overrightarrow{AC} \cdot \overrightarrow{BC} = (4)(1) + (-4)(-13) + (-4)(3) = 4 + 52 - 12 = 44$$ 5. **Calculate magnitudes:** $$|\overrightarrow{AC}| = \sqrt{4^2 + (-4)^2 + (-4)^2} = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3}$$ $$|\overrightarrow{BC}| = \sqrt{1^2 + (-13)^2 + 3^2} = \sqrt{1 + 169 + 9} = \sqrt{179}$$ 6. **Calculate cosine of angle:** $$\cos \theta = \frac{44}{4\sqrt{3} \times \sqrt{179}} = \frac{44}{4 \sqrt{537}} = \frac{\cancel{44}}{\cancel{4} \sqrt{537}} \times \frac{1}{1} = \frac{11}{\sqrt{537}}$$ 7. **Calculate angle $\theta$:** $$\theta = \cos^{-1} \left( \frac{11}{\sqrt{537}} \right)$$ Using a calculator, $$\theta \approx \cos^{-1}(0.474) \approx 61.7^\circ$$ **Final answer:** Vectors: $$\overrightarrow{AC} = (4, -4, -4), \quad \overrightarrow{BC} = (1, -13, 3)$$ Angle: $$\angle ACB \approx 61.7^\circ$$