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(C) Given the system of equations:$1) 2x + 3y - 2z = -4$$2) 3x - 4y + 3z = -5$$3) kx - 2y + z = -3$Since $x = \alpha = -2$, we substitute $x = -2$ int

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$\left| \begin{array}{ll} 3 & 5 \ 1 & 2 \end{array} ight|$

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(C) Given the system of equations:$1) 2x + 3y - 2z = -4$$2) 3x - 4y + 3z = -5$$3) kx - 2y + z = -3$Since $x = \alpha = -2$, we substitute $x = -2$ into the equations:$2(-2) + 3y - 2z = -4 \Rightarrow -4 + 3y - 2z = -4 \Rightarrow 3y - 2z = 0 \Rightarrow 2z = 3y \Rightarrow z = \frac{3}{2}y$$3(-2) - 4y + 3z = -5 \Rightarrow -6 - 4y + 3z = -5 \Rightarrow -4y + 3z = 1$Substitute $z = \frac{3}{2}y$ into the second equation:$-4y + 3(\frac{3}{2}y) = 1 \Rightarrow -4y + \frac{9}{2}y = 1 \Rightarrow \frac{1}{2}y = 1 \Rightarrow y = 2$Then $z = \frac{3}{2}(2) = 3$.Now substitute $x = -2, y = 2, z = 3$ into the third equation:$k(-2) - 2(2) + 3 = -3$$-2k - 4 + 3 = -3$$-2k - 1 = -3$$-2k = -2 \Rightarrow k = 1$.Evaluating the options:Option $C$: $\left| \begin{array}{ll} 3 & 5 \ 1 & 2 \end{array} ight| = (3 	imes 2) - (5 	imes 1) = 6 - 5 = 1$.Thus, $k = 1$ matches option $C$.

Let $x = \alpha, y = \beta, z = \gamma$ be the unique solution of the system of simultaneous linear equations $2x + 3y - 2z + 4 = 0$, $3x - 4y + 3z + 5 = 0$, and $kx - 2y + z + 3 = 0$. If $\alpha = -2$, then $k =$

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