Given: \((b + c - a)x = (c + a - b)y = (a + b - c)z = 2\) \[ \Rightarrow \frac{1}{(b + c - a)x} = \frac{1}{(c + a - b)y} = \frac{1}{(a + b - c)z} = \frac{1}{2} \] So: \[ \frac{1}{x} = \frac{b + c - a}{2}, \quad \frac{1}{y} = \frac{c + a - b}{2}, \quad \frac{1}{z} = \frac{a + b - c}{2} \] Now: \[ \frac{1}{x} + \frac{1}{y} = \frac{b + c - a}{2} + \frac{c + a - b}{2} = \frac{2c}{2} = c \] \[ \frac{1}{y} + \frac{1}{z} = \frac{c + a - b}{2} + \frac{a + b - c}{2} = \frac{2a}{2} = a \] \[ \frac{1}{z} + \frac{1}{x} = \frac{a + b - c}{2} + \frac{b + c - a}{2} = \frac{2b}{2} = b \] Therefore: \[ \left(\frac{1}{x} + \frac{1}{y}\right)\left(\frac{1}{y} + \frac{1}{z}\right)\left(\frac{1}{z} + \frac{1}{x}\right) = c \cdot a \cdot b = abc \quad \text{(Proved)} \]