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# If the system of linear equations $x+2ay+az=0,x+3by+bz=0,x+4cy+cz=0$ has a non-zero solution the a,b,c

$\begin{array}{1 1}(a)\;Satisfy \;a+2b+3c=0\\(b)\;are\;in\;H.P\\(c)\;are\;in\;G.P\\(d)\;are\;in\;H.P\end{array}$

For homogeneous system of equations to have non zero solution $\Delta=0$
$\begin{vmatrix}1 &2a&a\\1 &3b&b\\1&4c&c\end{vmatrix}=0$
Apply $C_2\rightarrow C_2-2C_3$
$\begin{vmatrix}1&0&a\\1&b&b\\1&2c&c\end{vmatrix}$
Apply $R_3\rightarrow R_3-R_2$
$R_2\rightarrow R_2-R_1$
$\begin{vmatrix}1&0&a\\0&b&b-a\\0&2c-b&c-b\end{vmatrix}$=0
$b(c-b)-(b-a)(2c-b)=0$
On simplification we get,
$\large\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
$\Rightarrow a,b,c$ are in harmonic progression.
hence (a) is the correct answer.