answer:Steps to solve the problem: 1. **Identify Initial Conditions:** In triangle ABC, we are given: [ AB = BC = 2, quad angle ABC = 120^circ ] Point P satisfies: [ PD = DA, quad PB = BA ] 2. **Construct Congruent Triangles:** By the given conditions, the triangles triangle ABD and triangle PBD are congruent. This implies: [ mathrm{d}(A, BD) = h quad text{and} quad mathrm{d}(P, ABC) le mathrm{d}(P, BD) = h ] The equality holds when the plane PBD is perpendicular to the plane ABC. 3. **Define Variables:** Let the length AD = x. Hence: [ x in (0, 2sqrt{3}) ] 4. **Calculate Length BD:** Using the Law of Cosines in triangle ABD: [ BD = sqrt{2^2 + x^2 - 2 cdot 2 cdot x cdot cos(30^circ)} ] Simplifying: [ BD = sqrt{x^2 - 2sqrt{3} cdot x + 4} ] 5. **Calculate height h:** The area S_{triangle ABD} can be written as: [ S_{triangle ABD} = frac{1}{2} cdot AB cdot AD cdot sin(30^circ) = x cdot frac{1}{2} ] Therefore, the height is: [ h = frac{2 cdot S_{triangle ABD}}{BD} = frac{x}{sqrt{x^2 - 2sqrt{3} cdot x + 4}} ] 6. **Find the Volume V_{P-BCD}:** Calculating the volume: [ V_{P-BCD} le frac{1}{3} cdot S_{CBD} cdot h ] Where: [ S_{CBD} = frac{1}{2} cdot BC cdot BD cdot sin(30^circ) = frac{1}{2} cdot 2 cdot BD cdot frac{1}{2} = frac{BD}{2} ] 7. **Perform Subsituion:** Substitute BD and h: [ V_{P-BCD} le frac{1}{3} cdot frac{BD}{2} cdot frac{x}{sqrt{x^2-2sqrt{3} cdot x + 4}} = frac{1}{6} cdot frac{x cdot BD}{sqrt{x^2-2sqrt{3} cdot x + 4}} ] Since: [ BD = sqrt{x^2 - 2sqrt{3} cdot x + 4}, quad text{we get:} ] [ V_{P-BCD} le frac{1}{6} cdot frac{x cdot (sqrt{x^2 - 2 sqrt{3} cdot x + 4})}{sqrt{x^2 - 2 sqrt{3} cdot x + 4}} = frac{1}{6} cdot x ] 8. **Transform Variable for Maximum Volume:** Let: [ x(2sqrt{3} - x) = t quad text{then} quad t in (0, 3) ] Substitute x: [ V_{P-BCD} le frac{1}{6} cdot frac{t}{sqrt{4 - t}} ] 9. **Maximize V_{P-BCD}:** To find maximum volume: [ V_{P-BCD} le frac{1}{6} cdot frac{3}{sqrt{4-3}} = frac{1}{2} ] 10. **Conclusion:** The maximum volume of the tetrahedron PBCD is: [ boxed{frac{1}{2}} ]

answer:Solution: (1) Since forall x in mathbb{R}, x^{2}-ax+a > 0 therefore Delta =a^{2}-4a < 0, solving this yields 0 < a < 4 (2) Since the equation dfrac {x^{2}}{a^{2}+12}- dfrac {y^{2}}{4-a^{2}}=1 represents a hyperbola therefore 4-a^{2} > 0, solving this yields -2 < a < 2 Since p is a false proposition, and q is a true proposition therefore begin{cases} aleqslant 0 text{ or } ageqslant 4 end{cases} within -2 < a < 2, therefore -2 < aleqslant 0. Thus, the final answers are: (1) The range of a when p is true is boxed{0 < a < 4}. (2) The range of a when p is false and q is true is boxed{-2 < a leqslant 0}.

answer:Given that the first term of the geometric sequence is a_1 = 3 and the common ratio is q, we have the first three terms as 3, 3q, 3q^2. The sum of these first three terms is expressed as: S_3 = a_1 + a_1q + a_1q^2 = 3 + 3q + 3q^2 According to the given information, S_3 = 21. Plugging the values into the equation gives us: 3 + 3q + 3q^2 = 21 Divide each term by 3 to simplify the equation: 1 + q + q^2 = 7 Rearrange the equation by moving all terms to one side: q^2 + q + 1 - 7 = 0 This simplifies to: q^2 + q - 6 = 0 Factor the quadratic equation: (q + 3)(q - 2) = 0 From the factorization, we get two possible solutions for q: q = -3 quad text{or} quad q = 2 However, since we're given that all terms of the sequence are positive numbers, q cannot be negative. Therefore, we discard q = -3. Thus, the value of q must be: boxed{q = 2}

answer:Given the operation heartsuit, defined by x heartsuit y = 4x + 6y, we need to find the value of 3 heartsuit 8. 1. Substitute x = 3 and y = 8 into the operation: [ 3 heartsuit 8 = 4(3) + 6(8) ] 2. Calculate the expression: [ 4(3) + 6(8) = 12 + 48 = 60 ] 3. Thus, the value of 3 heartsuit 8 is: [ boxed{60} ]