If the lines `a x+y+1=0,x+b y+1=0,` and `x+y+c=0(a , b , c` being distinct and different from 1) are concurrent, then `(1/(1-a))+(1/(1-b))+(1/(1-c))=` 0 (b) 1 `1/((a+b+c))` (d) none of these

If the lines ax+y+1=0,x+by+1=0 and x+y+c=0(a,b, being distinct and different from 1) are concurrent,then prove that (1)/(1-a)+(1)/(1-b)+(1)/(1-c)=1

If the lines ax+y+1=0, x+by+1=0 and x+y+c=0 (a,b and c being distinct and different from 1) are concurrent the value of (a)/(a-1)+(b)/(b-1)+(c)/(c-1) is

If the lines ax+y+1=0 , x+by+1=0 and x+y+c=0 , ( a , b , c being distinct and different from 1 ) are concurrent, then 1/(1-a)+1/(1-b)+1/(1-c)= (A)    0    (B)    1    (C)    1/(a+b+c)    (D)   None of these

If the lines ax+12y+1=0 bx+13y+1=0 and cx+14y+1=0 are concurrent then a,b,c are in

If the lines 2x-ay+1=0,3x-by+1=0,4x-cy+1=0 are concurrent,then a,b,c are in

If the lines ax+y+1=0,x+by+1=0 and x+y+c=0 are concurrent,prove that (1)/(1-a)+(1)/(1-b)+(1)/(1-c)=1

if the system of equation ax+y+z=0,x+by=z=0, and x+y+cz=0(a,b,c!=1) has a nontrival solution,then the value of (1)/(1-a)+(1)/(1-b)+(1)/(1-c)is:

If the lines ax+y+1=0,x+by+1=0 and x+y+c=0 are concurrent (a!=b!=c!=1), prove that (1)/(1-a)+(1)/(1-b)+(1)/(1-c)=1