If the system of linear equations 2x + 2ay + | vedclass

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(D) For a system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.The determinant is given by:$\

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$\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$

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(D) For a system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.The determinant is given by:$\begin{vmatrix} 2 & 2a & a \ 2 & 3b & b \ 2 & 4c & c \end{vmatrix} = 0$Taking $2$ common from the first column:$2 \begin{vmatrix} 1 & 2a & a \ 1 & 3b & b \ 1 & 4c & c \end{vmatrix} = 0$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$2 \begin{vmatrix} 1 & 2a & a \ 0 & 3b-2a & b-a \ 0 & 4c-2a & c-a \end{vmatrix} = 0$Expanding along the first column:$(3b-2a)(c-a) - (b-a)(4c-2a) = 0$Expanding the terms:$(3bc - 3ab - 2ac + 2a^2) - (4bc - 2ab - 4ac + 2a^2) = 0$$3bc - 3ab - 2ac + 2a^2 - 4bc + 2ab + 4ac - 2a^2 = 0$$-bc - ab + 2ac = 0$$2ac = ab + bc$Dividing both sides by $abc$ (since $a, b, c 
eq 0$):$\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$This condition implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$

If the system of linear equations $2x + 2ay + az = 0$,$2x + 3by + bz = 0$,and $2x + 4cy + cz = 0$,where $a, b, c \in R$ are non-zero and distinct,has a non-zero solution,then:

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