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

$\begin{array}{1 1}(A)\;\text{are in G.P}&(B)\;\text{are in H.P}\\(C)\;\text{satisfy a+2b+3c=0}&(D)\;\text{are in A.P}\end{array}$

$\begin{vmatrix}1&0&a\\1&b&b\\1&2c&c\end{vmatrix}=0$
On expanding along $R_1$ we get
$1(bc-2bc)+0(c-b)+a(2c-b)=0$
$\Rightarrow 2ac=ab+bc$
$b=\large\frac{2ac}{a+c}$
$\Rightarrow a,b,c$ in H.P