If \( x=2, y=3 \) and \( z=5 \), evaluate the following expressions. \( \begin{array}{llll}\text { a) } x+y-z & \text { b) } x-2 y+z & \text { c) } 5(x+y-3 z) & \text { d) } 3(z-x-y) \\ \begin{array}{llll}\text { e) } x^{2}+y^{2}-z^{2} & \text { f) }(2 x-y+z)^{3} & \text { g) }(2 x-4 y+3 z)^{2} & \text { h) } \frac{x^{2}-y^{2}+z^{3}}{x y z} \\ \text { i) } \frac{(x-y+z)^{2}}{x^{3}} & \text { j) } \frac{x}{y}+\frac{y}{z} & \text { k) } \frac{x}{y}-\frac{y}{z} & \text { 1) }(2 x-y)^{2}+z\end{array}\end{array} \).
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Let's evaluate the given expressions one by one using \( x = 2, y = 3, z = 5 \). a) \( x + y - z = 2 + 3 - 5 = 0 \) b) \( x - 2y + z = 2 - 2 \cdot 3 + 5 = 2 - 6 + 5 = 1 \) c) \( 5(x + y - 3z) = 5(2 + 3 - 3 \cdot 5) = 5(5 - 15) = 5(-10) = -50 \) d) \( 3(z - x - y) = 3(5 - 2 - 3) = 3(5 - 5) = 3(0) = 0 \) e) \( x^2 + y^2 - z^2 = 2^2 + 3^2 - 5^2 = 4 + 9 - 25 = 13 - 25 = -12 \) f) \( (2x - y + z)^3 = (2 \cdot 2 - 3 + 5)^3 = (4 - 3 + 5)^3 = (6)^3 = 216 \) g) \( (2x - 4y + 3z)^2 = (2 \cdot 2 - 4 \cdot 3 + 3 \cdot 5)^2 = (4 - 12 + 15)^2 = (7)^2 = 49 \) h) \( \frac{x^2 - y^2 + z^3}{xyz} = \frac{2^2 - 3^2 + 5^3}{2 \cdot 3 \cdot 5} = \frac{4 - 9 + 125}{30} = \frac{120}{30} = 4 \) i) \( \frac{(x - y + z)^2}{x^3} = \frac{(2 - 3 + 5)^2}{2^3} = \frac{(4)^2}{8} = \frac{16}{8} = 2 \) j) \( \frac{x}{y} + \frac{y}{z} = \frac{2}{3} + \frac{3}{5} = \frac{10}{15} + \frac{9}{15} = \frac{19}{15} \) k) \( \frac{x}{y} - \frac{y}{z} = \frac{2}{3} - \frac{3}{5} = \frac{10}{15} - \frac{9}{15} = \frac{1}{15} \) 1) \( (2x - y)^2 + z = (2 \cdot 2 - 3)^2 + 5 = (4 - 3)^2 + 5 = 1^2 + 5 = 1 + 5 = 6 \) Here are the final evaluated results: a) 0 b) 1 c) -50 d) 0 e) -12 f) 216 g) 49 h) 4 i) 2 j) \(\frac{19}{15}\) k) \(\frac{1}{15}\) 1) 6