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(C) For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the augmented matr

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$q-p=3$

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(C) For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the augmented matrix must be consistent.The coefficient matrix is $A = \begin{bmatrix} 1 & -2 & 1 \ 2 & 3 & 1 \ 1 & 2 & p \end{bmatrix}$.Setting $|A| = 0$:$1(3p - 2) - (-2)(2p - 1) + 1(4 - 3) = 0$$3p - 2 + 4p - 2 + 1 = 0$$7p - 3 = 0 \implies p = \frac{3}{7}$.For infinitely many solutions, the augmented matrix $[A|B]$ must have rank less than $3$. Using row operations on $[A|B] = \begin{bmatrix} 1 & -2 & 1 & | & 0 \ 2 & 3 & 1 & | & 6 \ 1 & 2 & 3/7 & | & q \end{bmatrix}$:$R_2 	o R_2 - 2R_1 \implies \begin{bmatrix} 1 & -2 & 1 & | & 0 \ 0 & 7 & -1 & | & 6 \ 1 & 2 & 3/7 & | & q \end{bmatrix}$$R_3 	o R_3 - R_1 \implies \begin{bmatrix} 1 & -2 & 1 & | & 0 \ 0 & 7 & -1 & | & 6 \ 0 & 4 & -4/7 & | & q \end{bmatrix}$$R_3 	o 7R_3 - 4R_2 \implies 0 = 7q - 24 \implies q = \frac{24}{7}$.Now, $pq = (\frac{3}{7})(\frac{24}{7}) = \frac{72}{49}$. Checking the options, there might be a typo in the provided options. Re-evaluating: $p = 3/7$ and $q = 24/7$. Thus $q-p = 21/7 = 3$. Option $C$ is correct.

If the system of simultaneous linear equations $x-2y+z=0$, $2x+3y+z=6$, and $x+2y+pz=q$ has infinitely many solutions, then:

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