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(D) Given the system of linear equations:$2x-3y+5z=12$ $(1)$$5x+2y+3z=11$ $(2)$$x+2y-3z=-3$ $(3)$First, calculate the determinant $D$ of the coefficie

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$2$

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(D) Given the system of linear equations:$2x-3y+5z=12$ $(1)$$5x+2y+3z=11$ $(2)$$x+2y-3z=-3$ $(3)$First, calculate the determinant $D$ of the coefficient matrix:$D = \begin{vmatrix} 2 & -3 & 5 \ 5 & 2 & 3 \ 1 & 2 & -3 \end{vmatrix} = 2(-6-6) + 3(-15-3) + 5(10-2) = 2(-12) + 3(-18) + 5(8) = -24 - 54 + 40 = -38$.Now, calculate $D_1, D_2, D_3$ using Cramer's Rule:$D_1 = \begin{vmatrix} 12 & -3 & 5 \ 11 & 2 & 3 \ -3 & 2 & -3 \end{vmatrix} = 12(-6-6) + 3(-33+9) + 5(22+6) = 12(-12) + 3(-24) + 5(28) = -144 - 72 + 140 = -76$.$D_2 = \begin{vmatrix} 2 & 12 & 5 \ 5 & 11 & 3 \ 1 & -3 & -3 \end{vmatrix} = 2(-33+9) - 12(-15-3) + 5(-15-11) = 2(-24) - 12(-18) + 5(-26) = -48 + 216 - 130 = 38$.$D_3 = \begin{vmatrix} 2 & -3 & 12 \ 5 & 2 & 11 \ 1 & 2 & -3 \end{vmatrix} = 2(-6-22) + 3(-15-11) + 12(10-2) = 2(-28) + 3(-26) + 12(8) = -56 - 78 + 96 = -38$.Solving for $\alpha, \beta, \gamma$:$\alpha = \frac{D_1}{D} = \frac{-76}{-38} = 2$.$\beta = \frac{D_2}{D} = \frac{38}{-38} = -1$.$\gamma = \frac{D_3}{D} = \frac{-38}{-38} = 1$.Finally, calculate $2\alpha + 5\beta + 3\gamma$:$2(2) + 5(-1) + 3(1) = 4 - 5 + 3 = 2$.

If $x=\alpha, y=\beta, z=\gamma$ is the unique solution of the system of linear equations $2x-3y+5z=12$, $5x+2y+3z=11$, and $x+2y-3z=-3$, then $2\alpha+5\beta+3\gamma=$

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