`-1 = (6k-3)/ (k+1)`

If the lines (x-1)/-3 = (y-2)/(2k) = (z-3)/2 and (x-1)/(3k) = (y-1)/1 = (z-6)/-5 are perpendicular, then find the value of k.

Find the value of k if the straight lines y-kx+4=0,(6k-3)x-(8k-1)y-6=0 are perpendicular.

Pairs of numbers of the form (6k-1) and (6k+1) are twin primes.Is this statement true?

Pairs of numbers of the form (6k-1) and (6k+1) are twin primes. Is this statement true ?

[(k (k + 1)) / (2)] ^ (2) + (k + 1) ^ (3) = [((k + 1) (k + 2)) / (2) +1] ^ (2)

If the lines (x-1)/(-3) =(y-2)/(2k) =(z-3)/2 and (x-1)/(3k)=(y-1)/1=(z-6)/(-5) are perpendicular , find the value of k .

If the lines (x-1)/-3 = (y-2)/2k = (z-3)/2 and (x-1)/3k = (y-1)/1 = (z-6)/-5 are perpendicular, find the value of k

If the lines (x -1)/(-3) = (x - 2)/(2k) = (z - 3)/(2) and (x - 1)/(3k ) = (y -1)/(1) = (z - 6)/(-5) are perpendicular to each other, then the value of k is ,

If cos3x+sin(2x-(7pi)/6)=-2 , then x is equal to (k in Z) pi/3(6k+1) (b) pi/3(6k-1) pi/3(2k+1) (d) none of these