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(B) Given the system of equations: $3x - 2y + z = 5k$ $2x + 3y - 2z = -5k$ $x + 4y + 3z = k$ First,calculate the determinant $D$ of the coefficient ma

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$2$

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(B) Given the system of equations: $3x - 2y + z = 5k$ $2x + 3y - 2z = -5k$ $x + 4y + 3z = k$ First,calculate the determinant $D$ of the coefficient matrix: $D = \begin{vmatrix} 3 & -2 & 1 \ 2 & 3 & -2 \ 1 & 4 & 3 \end{vmatrix} = 3(9 + 8) + 2(6 + 2) + 1(8 - 3) = 3(17) + 2(8) + 1(5) = 51 + 16 + 5 = 72$. Next,calculate $D_3$ using Cramer's rule: $D_3 = \begin{vmatrix} 3 & -2 & 5k \ 2 & 3 & -5k \ 1 & 4 & k \end{vmatrix} = k \begin{vmatrix} 3 & -2 & 5 \ 2 & 3 & -5 \ 1 & 4 & 1 \end{vmatrix} = k [3(3 + 20) + 2(2 + 5) + 5(8 - 3)] = k [3(23) + 2(7) + 5(5)] = k [69 + 14 + 25] = 108k$. Since $z = \frac{D_3}{D} = 3$,we have: $\frac{108k}{72} = 3$ $\frac{3k}{2} = 3$ $3k = 6$ $k = 2$.

If the unique solution of the simultaneous linear equations $3x - 2y + z = 5k$,$2x + 3y - 2z = -5k$,and $x + 4y + 3z = k$ is $x = \alpha, y = \beta, z = 3$,then $k =$

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