1. **Problem statement:** We want to find points $P$ on the line $g$ defined by $\vec{r} = t \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$ such that the angle $\angle APB = 90^\circ$, where $A = (7,5,2)$ and $B = (-2,4,6)$. 2. **Understanding the problem:** The angle $\angle APB$ is the angle at point $P$ formed by points $A$ and $B$. For this angle to be $90^\circ$, the vectors $\overrightarrow{PA}$ and $\overrightarrow{PB}$ must be perpendicular. 3. **Key formula:** Two vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if their dot product is zero: $$\vec{u} \cdot \vec{v} = 0.$$ 4. **Express $P$ on the line $g$:** Since $P$ lies on $g$, its coordinates are $$P = (t, 2t, 0).$$ 5. **Vectors from $P$ to $A$ and $B$:** $$\overrightarrow{PA} = A - P = (7 - t, 5 - 2t, 2 - 0) = (7 - t, 5 - 2t, 2),$$ $$\overrightarrow{PB} = B - P = (-2 - t, 4 - 2t, 6 - 0) = (-2 - t, 4 - 2t, 6).$$ 6. **Set dot product to zero:** $$\overrightarrow{PA} \cdot \overrightarrow{PB} = 0.$$ Calculate the dot product: $$(7 - t)(-2 - t) + (5 - 2t)(4 - 2t) + 2 \cdot 6 = 0.$$ 7. **Expand and simplify:** $$(7)(-2) + 7(-t) + (-t)(-2) + (-t)(-t) + (5)(4) - 5(2t) - 2t(4) + (2t)(2t) + 12 = 0,$$ which is $$-14 - 7t + 2t + t^2 + 20 - 10t - 8t + 4t^2 + 12 = 0.$$ 8. **Combine like terms:** $$t^2 + 4t^2 + (-7t + 2t - 10t - 8t) + (-14 + 20 + 12) = 0,$$ $$5t^2 - 23t + 18 = 0.$$ 9. **Solve quadratic equation:** $$5t^2 - 23t + 18 = 0.$$ Use the quadratic formula: $$t = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 5 \cdot 18}}{2 \cdot 5} = \frac{23 \pm \sqrt{529 - 360}}{10} = \frac{23 \pm \sqrt{169}}{10} = \frac{23 \pm 13}{10}.$$ 10. **Calculate roots:** $$t_1 = \frac{23 + 13}{10} = \frac{36}{10} = 3.6,$$ $$t_2 = \frac{23 - 13}{10} = \frac{10}{10} = 1.$$ 11. **Find points $P$ on $g$:** For $t_1 = 3.6$: $$P_1 = (3.6, 7.2, 0).$$ For $t_2 = 1$: $$P_2 = (1, 2, 0).$$ **Final answer:** The points $P$ on the line $g$ where $\angle APB = 90^\circ$ are $$P_1 = (3.6, 7.2, 0) \quad \text{and} \quad P_2 = (1, 2, 0).$$