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. (b)`alpha` is root of `4a x^2+12 b x+9c=0` or(c) `a , b , c` are in G.P. (d)`a , b , c ,` are in G.P. only `a , b ,c` are in A.P.

Options

A. a, b, c are in H.P.
B. `alpha` is a root of `4ax^(2) + 12bx + 9c = 0 or , a, b, c` are in G.P.
C. a, b, c are in G.P. only
D. a, b, c are in A.P.

Answer

B

Explanation

We have
`|(a,b,2a alpha + 3b),(b,c,2b alpha + 3c),(2a alpha + 3b ,2b alpha + 3c,0)|=0`
Applying `C_(3) rarr C_(3) - 2 alpha C_(1) - 3C_(2)`, we get
`|(a,b,0),(b,c,0),(2a alpha + 3b,2b alpha + 3c,-2 alpha (2a alpha + 3b) -3 (2b alpha + 3c))| = 0`
`rArr - (ac -b^(2)) {2alpha (2a alpha + 3b) + 3 (2b alpha + 3c)} = 0`
`rArr (b^(2) - ac) {4a alpha^(2) + 12b alpha + 9c} = 0`
`rArr b^(2) - ac = 0, 4a alpha^(2) + 12b alpha + 9c = 0`
`rArr a, b, c` are in G.P. and `x = alpha` is a root of `4ax^(2) + 12bx + 9c = 0`