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(D) Given the system of linear equations:$1) 2x - 3y + 2z = -15$$2) 3x + y - z = -2$$3) x - 3y - 3z = -8$From equation $(2)$,we have $z = 3x + y + 2$.

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$\alpha + \beta + \gamma = 0$

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(D) Given the system of linear equations:$1) 2x - 3y + 2z = -15$$2) 3x + y - z = -2$$3) x - 3y - 3z = -8$From equation $(2)$,we have $z = 3x + y + 2$.Substitute $z$ into equation $(1)$:$2x - 3y + 2(3x + y + 2) = -15$$2x - 3y + 6x + 2y + 4 = -15$$8x - y = -19$  --- $(4)$Substitute $z$ into equation $(3)$:$x - 3y - 3(3x + y + 2) = -8$$x - 3y - 9x - 3y - 6 = -8$$-8x - 6y = -2$  --- $(5)$Adding $(4)$ and $(5)$:$(8x - y) + (-8x - 6y) = -19 - 2$$-7y = -21 \implies y = 3$Substitute $y = 3$ into $(4)$:$8x - 3 = -19$$8x = -16 \implies x = -2$Substitute $x = -2$ and $y = 3$ into $z = 3x + y + 2$:$z = 3(-2) + 3 + 2 = -6 + 5 = -1$Thus,$\alpha = -2, \beta = 3, \gamma = -1$.Checking the options:$\alpha + \beta + \gamma = -2 + 3 - 1 = 0$.

If the values of $x, y$ and $z$ which satisfy the equations $2x - 3y + 2z + 15 = 0$,$3x + y - z + 2 = 0$ and $x - 3y - 3z + 8 = 0$ simultaneously are $\alpha, \beta$ and $\gamma$ respectively,then:

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Let $a, b, c, d, e$ be five numbers satisfying the system of equations:
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