answer:A point on the line is expressed as [begin{pmatrix} x y z end{pmatrix} = begin{pmatrix} 3 2 -1 end{pmatrix} + t begin{pmatrix} 1 -3 2 end{pmatrix} = begin{pmatrix} 3 + t 2 - 3t -1 + 2t end{pmatrix}.] The vector from (1, -1, 2) to (3 + t, 2 - 3t, -1 + 2t) is: [begin{pmatrix} 2 + t 3 - 3t -3 + 2t end{pmatrix}.] For the point on the line that is closest to (1, -1, 2), this vector must be orthogonal to the direction vector of the line, which is begin{pmatrix} 1 -3 2 end{pmatrix}. Thus, [begin{pmatrix} 2 + t 3 - 3t -3 + 2t end{pmatrix} cdot begin{pmatrix} 1 -3 2 end{pmatrix} = 0.] Expanding and solving this dot product equation gives: [(2 + t)(1) + (3 - 3t)(-3) + (-3 + 2t)(2) = 0.] [2 + t - 9 + 9t - 6 + 4t = 0.] [14t - 3 = 0.] [t = frac{3}{14}.] Substitute t = frac{3}{14} back into the point equation: [begin{pmatrix} x y z end{pmatrix} = begin{pmatrix} 3 + frac{3}{14} 2 - 3 cdot frac{3}{14} -1 + 2 cdot frac{3}{14} end{pmatrix} = begin{pmatrix} frac{45}{14} frac{16}{14} -frac{1}{7} end{pmatrix}.] So the closest point is boxed{left( frac{45}{14}, frac{16}{14}, -frac{1}{7} right)}.

answer:First, let's find out the cost of one puppy. Since the puppies were three times more expensive than the parakeets, and one parakeet costs 10, one puppy would cost 3 times 10, which is 30. Next, let's find out the cost of one kitten. Since the parakeets were half as expensive as the kittens, and one parakeet costs 10, one kitten would cost 2 times 10, which is 20. Now, let's calculate the total cost for all the pets: - The cost for two puppies would be 2 puppies * 30/puppy = 60. - The cost for two kittens would be 2 kittens * 20/kitten = 40. - The cost for three parakeets would be 3 parakeets * 10/parakeet = 30. Adding all these costs together, the total cost to purchase all of the pets would be 60 (puppies) + 40 (kittens) + 30 (parakeets) = boxed{130} .

answer:Let the number of girls at LHS be g and the number of boys be b. Then, we have: 1. The total number of students: g + b = 1500 2. The number of festival attendees: frac{3}{4}g + frac{2}{5}b = 820 From the total number of students equation: [ b = 1500 - g ] Substitute b in the festival attendees equation: [ frac{3}{4}g + frac{2}{5}(1500 - g) = 820 ] Multiply through by 20 to clear fractions: [ 15g + 8(1500 - g) = 16400 ] [ 15g + 12000 - 8g = 16400 ] [ 7g = 4400 ] [ g = frac{4400}{7} approx 628.57 ] Since the number of girls must be an integer, we round g to 628 (noting that we should check if 629 also works). The number of girls attending the festival is: [ frac{3}{4} times 628 = 471 ] Thus, the number of girls who attend the festival is boxed{471}.

answer:1. First, note that for a, b, c to be the sides of a triangle and their lengths not exceeding 2pi, we have 0 < a, b, c < pi. Consequently, sin a, sin b, and sin c are all positive real numbers since the sine function is positive in the interval (0, pi). Additionally, the cosine values for a, b, c will lie within the interval (-1, 1), i.e.: |cos a| < 1, quad |cos b| < 1, quad |cos c| < 1. 2. Without loss of generality, assume a leq b leq c. Let's first consider the case where a = frac{pi}{2}, then both b and c must also be equal to frac{pi}{2}. In this specific configuration, we observe that: sin a = 1, quad sin b = 1, quad sin c = 1. Therefore, the inequality sin a + sin b > sin c holds since: 1 + 1 = 2 > 1. 3. Next, consider the more general case where a neq frac{pi}{2}. We need to explore possible configurations: **Case 1: a + b + c = 2pi:** By the sine addition formula and properties of the sine function: sin c = sin(2pi - a - b) = -sin(a + b). Using properties of trigonometric functions: sin(a + b) = sin a cos b + cos a sin b. Therefore, sin c = -sin(a + b) = -(sin a cos b + cos a sin b). Taking absolute values and considering |cos a| < 1, |cos b| < 1: |sin c| = |sin(a + b)| leq |sin a cos b| + |cos a sin b| < sin a + sin b, which implies that: sin a + sin b > sin c. 4. **Case 2: a + b + c < 2pi:** To prove the triangle formation with sin values, we utilize another geometric argument. Assume a triangle with sides formed as: - OH is a perpendicular from point Q to the side PR. - Let OQ = OP = OR = 1, and the angles be formed as angle QOR = a, angle QOP = b, and angle POR = c. Applying the Law of Sines and trigonometric identities: sin a = sqrt{sin^2 varphi + cos^2 varphi sin^2 theta} = sqrt{sin^2 theta + sin^2 varphi cos^2 theta} geq |sin theta|, similarly for sin b: sin b geq |sin(c-theta)|. 5. To conclude this case, add the two inequalities and observe: sin a + sin b geq |sin theta| + |sin(c - theta)| geq |sin((c - theta) + theta)| = |sin c| = sin c. Thus, we've shown that in both configurations, the addition of the inequalities for the sine values satisfies the necessary conditions for forming a triangle. 6. Following these explorations and considerations, we deduce the relationships: sin a + sin b > sin c, quad sin b + sin c > sin a, quad sin a + sin c > sin b. # Conclusion: We have verified using inequalities, geometric arguments, and standard trigonometric properties that the condition holds. Therefore, the sides sin a, sin b, sin c can indeed form a triangle. blacksquare