answer:1. **Find the vector between the two points** (0, 2, 3) and (2, 0, 3): [ vec{v} = (2-0, 0-2, 3-3) = (2, -2, 0). ] 2. **Normal vector of the given plane** 3x - y + 2z = 7 is: [ vec{n} = (3, -1, 2). ] 3. **Calculate the cross product** of vec{v} and vec{n} to find the normal vector of the required plane: [ vec{v} times vec{n} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 2 & -2 & 0 3 & -1 & 2 end{vmatrix} = mathbf{i}(-4) + mathbf{j}(4) + mathbf{k}(-2) = (-4, 4, -2). ] Scaling to simplify: [ vec{n}_p = (-2, 2, -1). ] 4. **Derive the plane equation**. Use point (0, 2, 3): [ -2x + 2y - z + D = 0 quad text{Substituting } (0, 2, 3): ] [ -2 cdot 0 + 2 cdot 2 - 3 + D = 0 Rightarrow D = 1. ] So, the equation is: [ boxed{-2x + 2y - z + 1 = 0}. ] By multiplying by -1 for the positive x coefficient: [ boxed{2x - 2y + z - 1 = 0}. ]

answer:From the hyperbola frac{x^{2}}{4} - frac{y^{2}}{3} = 1, we can derive c = sqrt{4+3} = sqrt{7}, which gives us the foci (±sqrt{7}, 0). Given that the ellipse frac{x^{2}}{k} + frac{y^{2}}{9} = 1 and the hyperbola frac{x^{2}}{4} - frac{y^{2}}{3} = 1 share the same foci, we can find k using the relationship k - 9 = 7, which gives k = 16. Therefore, the length of the major axis of the ellipse is 2sqrt{16} = boxed{8}.

answer:1. **Identify the median**: The median of the set {m, m+3, m+8, n-1, n+3, 2n-2} is the average of the third and fourth smallest numbers. Since the median is n, we have: [ frac{(m+8) + (n-1)}{2} = n ] Simplifying this equation: [ m + 8 + n - 1 = 2n implies m + 7 = n ] 2. **Identify the mean**: The mean of the set is also given as n. Therefore, the sum of all elements divided by 6 must equal n: [ frac{m + (m+3) + (m+8) + (n-1) + (n+3) + (2n-2)}{6} = n ] Simplifying the sum: [ frac{3m + 3n + 11}{6} = n ] Multiplying through by 6: [ 3m + 3n + 11 = 6n implies 3m + 11 = 3n ] Substituting n = m + 7 from step 1: [ 3m + 11 = 3(m + 7) implies 3m + 11 = 3m + 21 ] This simplifies to: [ 11 = 21 quad text{(which is incorrect, indicating a mistake in calculation or assumption)} ] 3. **Re-evaluate the mean calculation**: Let's re-calculate the mean using the correct substitution n = m + 7: [ frac{3m + (m+7) + (m+10) + 2(m+7)-2}{6} = m + 7 ] Simplifying: [ frac{7m + 22}{6} = m + 7 ] Multiplying through by 6: [ 7m + 22 = 6m + 42 implies m = 20 ] Substituting m = 20 into n = m + 7: [ n = 20 + 7 = 27 ] 4. **Calculate m+n**: [ m + n = 20 + 27 = 47 ] Conclusion: [ 47 ] The final answer is boxed{C} (47).

answer:1. Determine the side length and altitude of the equilateral triangle. Since ( triangle ABC ) is equilateral with side length ( AB = 10 ) cm, the altitude can be calculated as: [ text{Altitude} = frac{sqrt{3}}{2} times text{Side length} = frac{sqrt{3}}{2} times 10 = 5sqrt{3} , text{cm} ] 2. Since ( CD ) is both the altitude of the equilateral triangle and the diameter of the circle: [ text{Diameter of the circle} = 5sqrt{3} , text{cm} ] Therefore, the radius ( l ) of the circle is: [ l = frac{5sqrt{3}}{2} , text{cm} ] 3. Analyze the segmented regions formed by the circle and the triangle. Each shaded region corresponds to a circular sector minus an isosceles triangle. 4. Calculate the area of one circular sector. Each sector has a central angle of ( 120^circ ) (or ( frac{2pi}{3} ) radians): [ text{Area of sector} = frac{120^circ}{360^circ} times pi l^2 = frac{1}{3} pi l^2 ] 5. Calculate the area of one isosceles triangle within the sector. The area of an isosceles triangle with angle ( 120^circ ) is: [ text{Area of triangle} = frac{1}{2} l^2 sin 120^circ = frac{1}{2} l^2 cdot frac{sqrt{3}}{2} = frac{sqrt{3}}{4} l^2 ] 6. Determine the area of one shaded region. The shaded region's area is the area of the sector minus the area of the isosceles triangle: [ text{Area of one shaded region} = frac{1}{3} pi l^2 - frac{sqrt{3}}{4} l^2 = frac{4pi - 3sqrt{3}}{12} l^2 ] 7. Substitute ( l = frac{5sqrt{3}}{2} ) into the area formula: [ text{Area of one shaded region} = frac{4pi - 3sqrt{3}}{12} left(frac{5 sqrt{3}}{2}right)^2 = frac{4pi - 3sqrt{3}}{12} cdot frac{75}{4} = frac{(4pi - 3sqrt{3}) cdot 75}{48} = frac{300pi - 225sqrt{3}}{48} = frac{100pi - 75sqrt{3}}{16} , text{cm}^2 ] 8. Calculate the total area for the two shaded regions: [ text{Total shaded area} = 2 times frac{100 pi - 75 sqrt{3}}{16} = frac{100 pi - 75 sqrt{3}}{8} , text{cm}^2 ] # Conclusion: The area of the shaded region is [ boxed{frac{100 pi - 75 sqrt{3}}{8} , text{cm}^2} ]