Mark the correct alternative in the following:

If the determinant , then


expend the determinats


a[-(2bα+3c)2 ]-b[-(2bα+3c)(2aα+3b)]+ (2aα+3b)[b(2bα+3c)-c(2aα+3b)]=0


-a(2bα+3c)2 + b(2bα+3c)(2aα+3b)+(2aα+3b)[2b^2 α+3bc-3bc-2acα]=0


(2bα+3c) [-2abα-3ac+2abα+3b2 ]+ (2aα+3b)(2α)( b2 -ac)=0


(2bα+3c) [-3ac +3b2 ]+ (2aα+3b)(2α)( b2-ac)=0


(b2 – ac)[4aα2 + 12bα + ac] = 0=


CASE1(b2 -ac)=0


b2 =ac {abc are in Gp}


CASE2(4aα2 +12bα+ac)=0 {Whose one root is }

1