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 H.P.

The correct answer is option (4) :  are in H.P.

The given system of equations will have a non-zero solution, if

$\begin{vmatrix}1 & 2a & 3\\1 & 3b & b\\1 & 4c & c\end{vmatrix}=0$

$⇒\begin{vmatrix}1 & 2a & a\\0 & 3b-2a & b-a\\0 & 4c-2a & c-a\end{vmatrix}$   Applying $R_2→R_2-R_1\, R_3→R_3-R_1$

$⇒(3b-2a) (c-a) -(b-a) (4c-2a) = 0 $

$⇒(3bc-3ab-2ca+2a^2) - ( 4bc - 2ab - 4ca +2a^2 ) = 0 $

$⇒ -bc - ab + 2 ca = 0 $

$⇒ \frac{2}{b}=\frac{1}{a}+\frac{1}{c}⇒a,b,c $ are in H.P.