Solve `(k-2)(2k+3)(k-1) >0`

Solve : 2(k + 3) + 3 (k-4) = 24

If the points A(k,2k) B(2k,3k) C(3,1) are collinear then k = ......... .

The point of concurrence of the lines (3k+1) x-(2k+3)y+(9-k)=0 is

Number of values of k for which the lines k x+y+1=0 x+k y+2=0 x+y+k=0 are concurrent is 3 (2) 2 (3) 1 (4) 0

Number of values of k for which the lines k x+y+1=0 x+k y+2=0 x+y+k=0 are concurrent is 3 (2) 2 (3) 1 (4) 0

Points (2k, 3k), (1, 0), (0, 1) and (0, 0) will be concyclic if the value of k is equal to-

A=[{:(1,0,k),(2, 1,3),(k,0,1):}] is invertible for