If abc neq 0 and the system of equations x+7a | vedclass

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(A) For the system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero. The system is: $x + 7ay

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harmonic progression

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(A) For the system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero. The system is: $x + 7ay + 2az = 0$ $x + 6by + 2bz = 0$ $x + 5cy + 2cz = 0$ The determinant $D$ is given by: $D = \begin{vmatrix} 1 & 7a & 2a \ 1 & 6b & 2b \ 1 & 5c & 2c \end{vmatrix} = 0$ Expanding the determinant along the first column: $1(12bc - 10bc) - 1(14ac - 10ac) + 1(14ab - 12ab) = 0$ $2bc - 4ac + 2ab = 0$ Dividing by $2$: $bc - 2ac + ab = 0$ $2ac = ab + bc$ Dividing both sides by $abc$ (since $abc 
eq 0$): $\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$ $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$ This condition implies that $a, b, c$ are in Harmonic Progression. Thus,option $A$ is correct.

If $abc \neq 0$ and the system of equations $x+7ay+2az=0$,$x+6by+2bz=0$,$x+5cy+2cz=0$ has a non-trivial solution,then $a, b, c$ are in

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