If `a+b+c!=0 ,ax+by+c=0, bx+cy+a=0, cx+ay+b=0` are concurrent then `(a^(2)+b^(2)+c^(2))/(ab+bc+ca)=`

If the lines ax+by+c=0,bx+cy+a=0 and cx+ay+b=0 be concurrent,then:

If a!=b!=c, lf ax+by+c=0,bx+cy+a=0 and cx+ay+b=0 are concurrent.Then the value of 2^(a^(2)b^(-1)c^(-1))2^(b^(2)c^(-1)a^(-1))2^(c^(2)a^(-1)b^(-1))

If the lines ax+by+c=0, bx+cy+a=0 and cx+ay+b=0 (a, b,c being distinct) are concurrent, then (A) a+b+c=0 (B) a+b+c=0 (C) ab+bc+ca=1 (D) ab+bc+ca=0

If a+b+c=0, then (a^(2))/(bc)+(b^(2))/(ca)+(c^(2))/(ab)=0 (b) 1(c)-1(d)3

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