answer:Let's follow the approach of Solution II to find the distance from the point (P) to the plane (CFH) in more detail. 1. **Place the cube in a coordinate system:** - The cube (ABCDEFGH) has unit-length edges. - Assign coordinates to the vertices such that (D) is at the origin: [D(0, 0, 0), quad A(1, 0, 0), quad B(1, 1, 0), quad C(0, 1, 0),] [H(0, 0, 1), quad G(1, 0, 1), quad F(1, 1, 1), quad E(0, 1, 1).] 2. **Find the third-point (P) on (BE):** - The coordinates of (B) are ( (1,1,0) ) and ( E ) are ( (0,1,1) ). The point (P) is one third of the way from (B) to (E), closer to (E). [ P = left(frac{2(1)+0}{3}, frac{2(1)+1}{3}, frac{2(0)+1}{3}right) = left(1, frac{2}{3}, frac{1}{3}right) ] 3. **Write the coordinates of vertices on the plane (CFH):** [ C(0, 1, 0), F(1, 1, 1), H(0, 0, 1) ] 4. **Determine the equation of the plane (CFH):** - The general form of a plane in 3D is: [ ax + by + cz + d = 0 ] - Use the points ( C(0, 1, 0) ), ( F(1, 1, 1) ), ( H(0, 0, 1) ): [ a(0) + b(1) + c(0) + d = 0 implies b + d = 0 Rightarrow d = -b ] [ a(1) + b(1) + c(1) + d = 0 Rightarrow a + b + c - b = 0 Rightarrow a + c = 0 Rightarrow c = -a ] [ a(0) + b(0) + c(1) + d = 0 implies c - b = 0 implies -a = b ] - Using ( a = k ), ( b = -k ), and ( c = -k ): [ kx - ky - kz + k = 0 ] - Dividing through by ( k ): [ x - y - z + 1 = 0 ] 5. **Compute the distance from point (P = left(1, frac{2}{3}, frac{1}{3}right)) to plane ( x - y - z + 1 = 0 ):** The distance (d) from a point ((x_1, y_1, z_1)) to the plane ( ax + by + cz + d = 0 ) is given by: [ d = frac{|ax_1 + by_1 + cz_1 + d|}{sqrt{a^2 + b^2 + c^2}} ] Substituting the values (a=1, b=-1, c=-1, d = 1): [ d = frac{|1(1) -1left(frac{2}{3}right) -1left(frac{1}{3}right) + 1|}{sqrt{1^2 + (-1)^2 + (-1)^2}} ] Simplify the numerator: [ = frac{|1 -frac{2}{3} -frac{1}{3} + 1|}{sqrt{1+1+1}} ] [ = frac{|1 -frac{3}{3} + 1|}{sqrt{3}} ] [ = frac{|1 - 1 + 1|}{sqrt{3}} ] [ = frac{|1|}{sqrt{3}} ] [ = frac{1}{sqrt{3}} = frac{sqrt{3}}{3} ] # Conclusion: (boxed{frac{sqrt{3}}{3}})

answer:Let lg log_{3}10=m, then lg lg 3=-lg log_{3}10=-m. Since f(lg log_{3}10)=5, and f(x)=asin x+b cdot 3x +4, we have f(m)=asin m+b cdot 3m +4=5, thus asin m+b cdot 3m =1. Therefore, f(-m)=-(asin m+b cdot 3m )+4, which equals -1+4, equals 3. Hence, the correct choice is boxed{C}. By setting lg log_{3}10=m, we find that lg lg 3=-lg log_{3}10=-m. From this, we can determine the value of f(lg lg 3). This problem tests the method of finding function values, which is a basic question. When solving, it is important to carefully read the problem and meticulously work through the solution.

answer:**Analysis** This problem examines the use of the sum of the first n terms of a sequence to find the general formula of the sequence, and the study of inequalities that always hold using the function of the sequence. **Solution** Given S_n=n^2+2n, When n=1, a_1=3; When n geqslant 2, a_n=S_n-S_{n-1}=n^2+2n-left[(n-1)^2+2(n-1)right]=2n+1, When n is even, b_n=-(2n+1)(2n+3); T_n=3 times 5 - 5 times 7 + 7 times 9 - 9 times 11 + cdots - (2n+1)(2n+3)=(-4)(5+9+cdots+2n+1)=-2n^2-6n, Then T_n geqslant tn^2 can be simplified to t leqslant left(-2- frac{6}{n}right)_{min}, Thus t leqslant -2- frac{6}{2}=-5; When n is odd, b_n=(2n+1)(2n+3), T_n=T_{n-1}+(2n+1)(2n+3)=2n^2+6n+7, So t leqslant 2+ frac{6}{n}+ frac{7}{n^2}, Then t leqslant 2, Therefore, t leqslant -5. Hence, the answer is boxed{(-infty, -5]}.

answer:We need to prove that for every positive integer n, there exist n consecutive positive integers such that none of them are a power of a prime number. 1. **Define the sequence of numbers**: We consider the set of n consecutive integers starting from 2 + M, where M will be chosen later. The sequence is: [ {2+M, 3+M, ldots, (n+1)+M} ] 2. **Condition for not being a prime power**: A positive integer is a power of a prime if it can be expressed as p^k where p is a prime number and k is an integer greater than or equal to 1. A number is not a prime power if it has at least two distinct prime factors. 3. **Choosing M**: To ensure none of the numbers in the sequence are prime powers, we select ( M ) such that: [ M = [(n+1)!]^2 ] Here, (n+1)! denotes the factorial of (n+1). 4. **Validity of the choice of M**: With ( M = [(n+1)!]^2 ), each of the numbers in our sequence can be written as: [ i + M quad text{for} quad i = 2, 3, ldots, (n+1) ] 5. **Ensuring numbers are not prime powers**: We verify the form of each number: [ i + M = i + [(n+1)!]^2 ] For each ( i ) in the range 2 leq i leq n+1, the factorial term (n+1)! includes all integers up to n+1, and thus each term ( i + M ) has the form: [ i + [(n+1)!]^2 = i + (k cdot i^2) quad text{where} quad k = left(frac{[(n+1)!]}{i}right)^2 ] Since ( i ) has been factored out and the resulting term ( 1 + k cdot i ) is coprime to ( left(1+kright) cdot i ), each number in our sequence is composed of at least two distinct prime factors. Thus, none of the numbers can be a prime power. In conclusion, we have shown that there exist n consecutive positive integers such that none of them are a power of a prime number by selecting M = [(n+1)!]^2. blacksquare