If the system of equations x + 2y + 3z = 4,x | vedclass

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(D) For a system of linear equations to have an infinite number of solutions,the determinant of the coefficient matrix $D$ must be $0$,and $D_x = D_y

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None of these

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(D) For a system of linear equations to have an infinite number of solutions,the determinant of the coefficient matrix $D$ must be $0$,and $D_x = D_y = D_z = 0$.The coefficient matrix is $A = \begin{bmatrix} 1 & 2 & 3 \ 1 & p & 2 \ 1 & 4 & \mu \end{bmatrix}$.Setting $D = 0$:$1(p\mu - 8) - 2(\mu - 2) + 3(4 - p) = 0$$p\mu - 8 - 2\mu + 4 + 12 - 3p = 0$$p\mu - 3p - 2\mu + 8 = 0$$(p - 2)(\mu - 3) = 2$ (This does not immediately yield a simple integer solution).Alternatively,using the augmented matrix $\begin{bmatrix} 1 & 2 & 3 & | & 4 \ 1 & p & 2 & | & 3 \ 1 & 4 & \mu & | & 3 \end{bmatrix}$.Subtracting row $1$ from row $2$ and row $3$:$R_2 	o R_2 - R_1: (0, p-2, -1, | -1)$$R_3 	o R_3 - R_1: (0, 2, \mu-3, | -1)$For infinite solutions,the rows must be proportional: $\frac{p-2}{2} = \frac{-1}{\mu-3} = \frac{-1}{-1} = 1$.Thus,$p-2 = 2 \implies p = 4$ and $\mu-3 = -1 \implies \mu = 2$.Checking the options,none match $p=4, \mu=2$.

List-$I$	List-$II$
$(A)$ $\lambda=8, \mu \neq 15$	$1$. Infinitely many solutions
$(B)$ $\lambda \neq 8, \mu \in R$	$2$. No solution
$(C)$ $\lambda=8, \mu=15$	$3$. Unique solution

If the system of equations $x + 2y + 3z = 4$,$x + py + 2z = 3$,and $x + 4y + \mu z = 3$ has an infinite number of solutions,then:

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