Given: \(\frac{ay - bx}{c} = \frac{cx - az}{b} = \frac{bz - cy}{a}\) ∴ \(\frac{c(ay - bx)}{c^2} = \frac{b(cx - az)}{b^2} = \frac{a(bz - cy)}{a^2}\) Now, combining all three expressions: \(\frac{c(ay - bx) + b(cx - az) + a(bz - cy)}{c^2 + b^2 + a^2}\) Using addition: \(= \frac{acy - bcx + bcx - abz + abz - acy}{a^2 + b^2 + c^2}\) \(= \frac{0}{a^2 + b^2 + c^2} = 0\) Therefore, \(\frac{ay - bx}{c} = 0\) ⇒ \(ay - bx = 0\) ⇒ \(ay = bx\) ⇒ \(\frac{y}{b} = \frac{x}{a}\) Also, \(\frac{cx - az}{b} = 0\) ⇒ \(cx = az\) ⇒ \(\frac{x}{a} = \frac{z}{c}\) ∴ \(\frac{x}{a} = \frac{y}{b} = \frac{z}{c}\) — proved.