\( \left. \begin{array} { l } { 3 x + y + 2 z = 18 } \\ { 2 x + 4 y - 3 z = 20 } \\ { x + y - 3 z = 12 } \\ { \Rightarrow x - y + z = ? } \end{array} \right. \)

To solve for \( x - y + z \), we first need to solve the system of equations you've listed. We have: 1. \( 3x + y + 2z = 18 \) 2. \( 2x + 4y - 3z = 20 \) 3. \( x + y - 3z = 12 \) Let's start by using the third equation to express \( y \): From \( x + y - 3z = 12 \): \[ y = 12 - x + 3z \] Next, substitute \( y \) into the first and second equations: Plugging into the first equation: \[ 3x + (12 - x + 3z) + 2z = 18 \] This simplifies to: \[ 2x + 5z + 12 = 18 \] Thus: \[ 2x + 5z = 6 \] (Equation 4) Now plug \( y = 12 - x + 3z \) into the second equation: \[ 2x + 4(12 - x + 3z) - 3z = 20 \] This gives: \[ 2x + 48 - 4x + 12z - 3z = 20 \] Simplifying this results in: \[ -2x + 9z + 48 = 20 \] So: \[ -2x + 9z = -28 \] Or: \[ 2x - 9z = 28 \] (Equation 5) Now we have two new equations (4 and 5): 1. \( 2x + 5z = 6 \) 2. \( 2x - 9z = 28 \) Now we can subtract these equations: \[ (2x - 9z) - (2x + 5z) = 28 - 6 \] This leads to: \[ -14z = 22 \] Thus: \[ z = -\frac{22}{14} = -\frac{11}{7} \] Now substitute \( z \) back into Equation 4 \( 2x + 5z = 6 \): \[ 2x + 5(-\frac{11}{7}) = 6 \] This simplifies to: \[ 2x - \frac{55}{7} = 6 \] Now multiply everything by 7 to eliminate the fraction: \[ 14x - 55 = 42 \] Solving for \( x \) gives: \[ 14x = 97 \] So: \[ x = \frac{97}{14} \] And finally, substitute \( x \) and \( z \) back into the expression for \( y \): \[ y = 12 - \frac{97}{14} + 3(-\frac{11}{7}) \] Calculating: \[ y = 12 - \frac{97}{14} - \frac{33}{7} = 12 - \frac{97}{14} - \frac{66}{14} \] This gives: \[ y = 12 - \frac{163}{14} = 12 - \frac{163}{14} = \frac{168 - 163}{14} = \frac{5}{14} \] Finally, we compute: \[ x - y + z = \frac{97}{14} - \frac{5}{14} - \frac{11}{7} \] Convert \( z \) to a denominator of 14: \[ z = -\frac{22}{14} \] So: \[ x - y + z = \frac{97}{14} - \frac{5}{14} - \frac{22}{14} = \frac{70}{14} = 5 \] Thus, \( x - y + z = 5 \).