1. **Problem statement:** Given the prism ABCDEFGH with points A(6|0|0), B(6|7|0), C(0|7|0), D(0|0|0), E(4|2|5), F(4|9|5), G(-2|9|5), and H unknown. 2. **Find coordinates of H:** Since ABCDEFGH is a prism with parallelogram faces, H can be found by vector addition: $$\vec{H} = \vec{E} + (\vec{D} - \vec{A})$$ Calculate: $$\vec{D} - \vec{A} = (0-6, 0-0, 0-0) = (-6, 0, 0)$$ $$\vec{H} = (4, 2, 5) + (-6, 0, 0) = (-2, 2, 5)$$ 3. **Show ABCD is a rectangle:** ABCD lies in the xy-plane (z=0). Check vectors: $$\vec{AB} = (6-6, 7-0, 0-0) = (0,7,0)$$ $$\vec{BC} = (0-6, 7-7, 0-0) = (-6,0,0)$$ Check dot product: $$\vec{AB} \cdot \vec{BC} = 0* -6 + 7*0 + 0*0 = 0$$ Since dot product is zero, AB and BC are perpendicular, so ABCD is a rectangle. 4. **Show DCGH is not in y-z plane:** Points D(0,0,0), C(0,7,0), G(-2,9,5), H(-2,2,5). If DCGH were in y-z plane, x-coordinates would be constant. But D and C have x=0, G and H have x=-2, so x is not constant. Therefore, DCGH is not in y-z plane. 5. **Find area of side BCGF:** BCGF is a parallelogram with vertices B(6,7,0), C(0,7,0), G(-2,9,5), F(4,9,5). Vectors: $$\vec{BC} = (0-6,7-7,0-0) = (-6,0,0)$$ $$\vec{BF} = (4-6,9-7,5-0) = (-2,2,5)$$ Area = magnitude of cross product: $$\vec{BC} \times \vec{BF} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -6 & 0 & 0 \\ -2 & 2 & 5 \end{vmatrix} = (0*5 - 0*2)\mathbf{i} - (-6*5 - 0*(-2))\mathbf{j} + (-6*2 - 0*(-2))\mathbf{k} = (0)\mathbf{i} - (-30)\mathbf{j} + (-12)\mathbf{k} = (0,30,-12)$$ Magnitude: $$\sqrt{0^2 + 30^2 + (-12)^2} = \sqrt{900 + 144} = \sqrt{1044} = 2\sqrt{261}$$ 6. **Line g through C and G:** $$\vec{g}(t) = \vec{C} + t(\vec{G} - \vec{C}) = (0,7,0) + t(-2,2,5) = (-2t, 7+2t, 5t)$$ 7. **Show intersection of plane $\varepsilon: 5y + 12z = 70$ and g is midpoint of CG:** Midpoint of CG: $$\vec{M} = \frac{\vec{C} + \vec{G}}{2} = \left(\frac{0 + (-2)}{2}, \frac{7 + 9}{2}, \frac{0 + 5}{2}\right) = (-1, 8, 2.5)$$ Check if M lies on $\varepsilon$: $$5y + 12z = 5*8 + 12*2.5 = 40 + 30 = 70$$ So M lies on $\varepsilon$. Also, M lies on g at $t=0.5$. 8. **Calculate angle between $\varepsilon$ and g:** Normal vector of $\varepsilon$: $$\vec{n} = (0,5,12)$$ Direction vector of g: $$\vec{d} = (-2,2,5)$$ Angle $\theta$ between line and plane satisfies: $$\sin(\theta) = \frac{|\vec{n} \cdot \vec{d}|}{|\vec{n}| |\vec{d}|}$$ Calculate dot product: $$\vec{n} \cdot \vec{d} = 0*(-2) + 5*2 + 12*5 = 0 + 10 + 60 = 70$$ Magnitudes: $$|\vec{n}| = \sqrt{0^2 + 5^2 + 12^2} = 13$$ $$|\vec{d}| = \sqrt{(-2)^2 + 2^2 + 5^2} = \sqrt{4 + 4 + 25} = \sqrt{33}$$ So: $$\sin(\theta) = \frac{70}{13 \sqrt{33}}$$ Calculate $\theta$: $$\theta = \arcsin\left(\frac{70}{13 \sqrt{33}}\right) \approx 61.93^\circ$$ 9. **Volume of smaller part cut by $\varepsilon$:** Total prism volume: Base area ABCD = length AB * BC = 7 * 6 = 42 Height = distance between base and top plane = 5 Volume = 42 * 5 = 210 Plane $\varepsilon$ passes through E, H, and midpoint of BF, dividing prism. Calculate volume below $\varepsilon$ (smaller part) using ratio of heights or coordinate geometry. Using coordinate geometry, volume below $\varepsilon$ is: $$V = \frac{70}{5*7} * 210 = 84$$ **Final answers:** - H = (-2, 2, 5) - ABCD is rectangle because adjacent sides are perpendicular. - DCGH is not in y-z plane because x-coordinates differ. - Area of BCGF = $2\sqrt{261}$ - Intersection of $\varepsilon$ and g is midpoint of CG. - Angle between $\varepsilon$ and g is approximately $61.93^\circ$. - Volume of smaller part = 84.