if the system of linear equations
`{:(x+2ay+az=0),(x+3by+bz=0),(x+4cy+cz=0):}`
has a non zero solution then a,b,c are in

Options

A. satisfy `a + 2b + 3c = 0`
B. are in A.P.
C. are in G.P.
D. are in H.P.

Answer

D

Explanation

The given system of equations will have a non-zero solution, if
`|(1,2a,a),(1,3b,b),(1,4c,c)| = 0`
`rArr |(1,2a,a),(0,3b - 2a,b - a),(0,4c - 2a,c -a)| "Applying " R_(2) rarr R_(2) - R_(1) R_(3) rarr R_(3) - R_(1)`
`rArr (3b - 2a) (c -a) - (b -a) (4c - 2a) = 0`
`rArr (3bc - 3ab - 2ca + 2a^(2)) - (4bc - 2ab - 4ca + 2a^(2)) = 0`
`rArr - bc - ab + 2ca = 0`
`rArr (2)/(b) - (1)/(a) + (1)/(c) rArr a, b, c` are in H.P.