If` a+b+c=n,a>0,b>0,c>0`, then `a^pb^qc^r` `<=`

If a x^2+b x+c=0,a ,b ,c in R has no real zeros, and if c 0 (c) a >0

If a x^2+b x+c=0,a ,b ,c in R has no real zeros, and if c 0 (c) a >0

If p+q+r=0 and |(pa,qb,rc),(qc,ra,pb),(rb,pc,qa)| = lambda |(a,b,c),(c,a,b),(b,c,a)| then lambda =

If p+q+r=0 and |(pa,qb,rc),(qc,ra,pb),(rb,pc,qa)| = lambda |(a,b,c),(c,a,b),(b,c,a)| then lambda =

If |0a b^2a c^2a^2b0b c^2a^2cc b^2 0|=2a^pb^q c^r ,t h e np+q+r+10 is equal to 27 (b) 18 (c) 19 (d) 8

If a!=b!=c\ a n d\ |{:(a, b, c), (a^2,b^2,c^2), (b+c, c+a, a+b):}|=0 then a+b+c=0 b. a b+b c+c a=0 c. a^2+b^2+c^2=a b+b c+c a d. a b c=0

If a!=b!=c\ a n d\ |{:(a, b, c), (a^2,b^2,c^2), (b+c, c+a, a+b):}|=0 then a+b+c=0 b. a b+b c+c a=0 c. a^2+b^2+c^2=a b+b c+c a d. a b c=0

If a+b+c=0, an^(2)+bn+c=0 and a+bn+cn^(2)=0 where n!=0, 1 , then prove that a=b=c=0

If |(a,-b,a-b-c),(-a,b,-a+b-c),(-a,-b,-a-b+c)| - kabc = 0 (a != 0, b != 0, c != 0) then what is the value of k ?