1. **Problem statement:** Find the area of the triangle with vertices $A(1,-2,2)$, $B(5,-6,2)$, and $C(1,3,-1)$, and find the height $h$ from point $B$ to the line segment $AC$. 2. **Formula for area of a triangle using vectors:** The area $S$ of a triangle with vertices $A$, $B$, and $C$ can be found using the cross product: $$S = \frac{1}{2} \| \overrightarrow{AB} \times \overrightarrow{AC} \|$$ 3. **Calculate vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:** $$\overrightarrow{AB} = B - A = (5-1, -6+2, 2-2) = (4, -4, 0)$$ $$\overrightarrow{AC} = C - A = (1-1, 3+2, -1-2) = (0, 5, -3)$$ 4. **Calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$:** $$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -4 & 0 \\ 0 & 5 & -3 \end{vmatrix} = \mathbf{i}((-4)(-3) - 0 \cdot 5) - \mathbf{j}(4(-3) - 0 \cdot 0) + \mathbf{k}(4 \cdot 5 - (-4) \cdot 0)$$ $$= \mathbf{i}(12) - \mathbf{j}(-12) + \mathbf{k}(20) = (12, 12, 20)$$ 5. **Calculate the magnitude of the cross product:** $$\| \overrightarrow{AB} \times \overrightarrow{AC} \| = \sqrt{12^2 + 12^2 + 20^2} = \sqrt{144 + 144 + 400} = \sqrt{688} = 2\sqrt{172}$$ 6. **Calculate the area $S$ of the triangle:** $$S = \frac{1}{2} \times 2\sqrt{172} = \sqrt{172} = 2\sqrt{43}$$ 7. **Find the length of $AC$ to calculate height $h$:** $$|\overrightarrow{AC}| = \sqrt{0^2 + 5^2 + (-3)^2} = \sqrt{0 + 25 + 9} = \sqrt{34}$$ 8. **Use the formula for height $h$ from base $AC$:** $$S = \frac{1}{2} \times |AC| \times h \implies h = \frac{2S}{|AC|} = \frac{2 \times 2\sqrt{43}}{\sqrt{34}} = \frac{4\sqrt{43}}{\sqrt{34}} = 4 \sqrt{\frac{43}{34}}$$ **Final answers:** - Area of the triangle: $2\sqrt{43}$ - Height from $B$ to $AC$: $4 \sqrt{\frac{43}{34}}$