If the determinant `|[a,b,2aalpha+3b],[b,c,2balpha+3c],[2aalpha+3b,2balpha+3c,0]|=0` then (a)`a , b , c` are in H.P. `alpha` is root of `4a x^2+12 b x+9c=0` or `a , b , c` are in G.P. `a , b , c ,` are in G.P. only `a , b ,c` are in A.P.

Explanation

`|(a, b , 2a alpha + 3b), (b,c,2b alpha + 3c),(2a alpha + 3b, 2b alpha +3c, 0)|= 0`

`a[-(2b alpha + 3c^2)] - b[-(2b alpha + 3c)(2a alpha + 3b)] + (2a alpha + 3b)[ b(2 b alpha + 3c) - c(2 a alpha + 3b)]= 0`

`=> -a(2b alpha + 3c)^2 + b(2b alpha + 3c)(2 a alpha + 3b) + (2a alpha + 3b)[ 2a^2 alpha + 3bc - 2ac alpha - 3bc] = 0`

`= -a(2b alpha + 3c)^2 + b(2b alpha + 3c)(2a alpha + 3b) + (2a alpha + 3b)[ 2 alpha (b^2 - ac] = 0`

`= (2b alpha + 3c)[ -2ab alpha - 3ac + 2ab alpha + 3b^2] + (2 a alpha + 3b)(2a) (b^2 - ac) = 0`

`= (2b alpha + 3c)(3b^2 - 3ac) + (2 alpha)(2a alpha + 3b)(b^2 - ac) = 0`

`= (b^2 -ac)[ 6b alpha + 9c + 4a alpha^2 + 6b alpha]`

`= (b^2- ac)[4a alpha^2 + 12 alpha b + 9c] = 0`

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