If the system of equations 2x+ 2ay + az= 0 2x+3by + bz =0 2x+ 4cy + c…

The correct answer is option (3) : $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

If the given system of equations have a non-trivial solution, then

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

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

Applying $R_2→R_2-R_1,R_3→R_3-R_1$

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

$⇒(3b-2a)(c-a) -(4c-2a) (b-a)=0$

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

$⇒-bc - ab + 2ca = 0 $

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