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(C) For a system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.The determinant is given b

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$a, b, c$ are in $HP$

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(C) For a system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.The determinant is given by:$\left|\begin{array}{ccc}1 & 4a & a \ 1 & 3b & b \ 1 & 2c & c\end{array}ight|=0$Expanding along the first column:$1(3bc - 2bc) - 1(4ac - 2ac) + 1(4ab - 3ab) = 0$Simplifying the expression:$(bc) - (2ac) + (ab) = 0$$bc + ab = 2ac$Dividing both sides by $abc$ (assuming $a, b, c 
eq 0$):$\frac{bc}{abc} + \frac{ab}{abc} = \frac{2ac}{abc}$$\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$This condition implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $AP$, which means $a, b, c$ are in $HP$.

If the following three linear equations have a non-trivial solution, then
$x+4ay+az=0$
$x+3by+bz=0$
$x+2cy+cz=0$

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If the following three linear equations have a non-trivial solution, then$x+4ay+az=0$$x+3by+bz=0$$x+2cy+cz=0$