1. **State the problem:** Find the point $P$ on line segment $AC$ such that the height $h$ from point $B$ to $AC$ is perpendicular to $AC$. 2. **Formula and rules:** The height from $B$ to $AC$ is the perpendicular distance from $B$ to the line through $A$ and $C$. Point $P$ lies on $AC$ and satisfies $\overrightarrow{BP} \perp \overrightarrow{AC}$. 3. **Find vector $\overrightarrow{AC}$:** $$\overrightarrow{AC} = C - A = (1 - 1, 3 - (-2), -1 - 2) = (0, 5, -3)$$ 4. **Parameterize point $P$ on $AC$:** $$P = A + t\overrightarrow{AC} = (1, -2, 2) + t(0, 5, -3) = (1, -2 + 5t, 2 - 3t)$$ 5. **Vector $\overrightarrow{BP}$:** $$\overrightarrow{BP} = P - B = (1 - 5, (-2 + 5t) - (-6), (2 - 3t) - 2) = (-4, 4 + 5t, -3t)$$ 6. **Perpendicularity condition:** $$\overrightarrow{BP} \cdot \overrightarrow{AC} = 0$$ Calculate the dot product: $$(-4)(0) + (4 + 5t)(5) + (-3t)(-3) = 0$$ $$0 + 20 + 25t + 9t = 0$$ $$20 + 34t = 0$$ 7. **Solve for $t$:** $$34t = -20$$ $$t = -\frac{20}{34} = -\frac{10}{17}$$ 8. **Find coordinates of $P$:** $$P = (1, -2 + 5(-\frac{10}{17}), 2 - 3(-\frac{10}{17})) = \left(1, -2 - \frac{50}{17}, 2 + \frac{30}{17}\right)$$ Simplify: $$y = -2 - \frac{50}{17} = -\frac{34}{17} - \frac{50}{17} = -\frac{84}{17}$$ $$z = 2 + \frac{30}{17} = \frac{34}{17} + \frac{30}{17} = \frac{64}{17}$$ So, $$P = \left(1, -\frac{84}{17}, \frac{64}{17}\right)$$ **Final answer:** $$P = \left(1, -\frac{84}{17}, \frac{64}{17}\right)$$