By eliminating a,b,c from the homogenous equations x=ab−c,y=bc−a,z=ca−b where a,b,c not all zero

x=ab−c⇒a=xb−xc⇒a−xb+xc=0→(1)y=bc−a⇒b=yc−ya⇒ya+b−yc=0→(2)z=ca−b⇒c=za−zb⇒za−zb−c=0→(3) By eliminating a.b,c from (1), (2), (3), we get 1−xxy1−yz−z−1=0⇒1(−1−yz)+x(−y+yz)+x(−yz−z)=0⇒−1−yz−xy+xyz−xyz−zx=0⇒xy+yz+zx=−1

By eliminating a,b,c from the homogenous equations x=ab−c,y=bc−a,z=ca−b where a,b,c not all zero

x=ab−c⇒a=xb−xc⇒a−xb+xc=0→(1)y=bc−a⇒b=yc−ya⇒ya+b−yc=0→(2)z=ca−b⇒c=za−zb⇒za−zb−c=0→(3) By eliminating a.b,c from (1), (2), (3), we get 1−xxy1−yz−z−1=0⇒1(−1−yz)+x(−y+yz)+x(−yz−z)=0⇒−1−yz−xy+xyz−xyz−zx=0⇒xy+yz+zx=−1

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By eliminating $a, b, c$ from the homogenous equations $x = \frac{a}{b - c}, y = \frac{b}{c - a}, z = \frac{c}{a - b}$ where $a, b, c$ not all zero

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$x + y + z = 1$

$x y + y z + z x = - 1$

$x y + y z + z x = 1$

$x + y + z = 0$

answer is B.

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Detailed Solution

$\begin{aligned} x = \frac{a}{b - c} \Rightarrow a = x b - x c \Rightarrow a - x b + x c = 0 \to (1) \\ y = \frac{b}{c - a} \Rightarrow b = y c - y a \Rightarrow y a + b - y c = 0 \to (2) \\ z = \frac{c}{a - b} \Rightarrow c = z a - z b \Rightarrow z a - z b - c = 0 \to (3) \end{aligned}$

$By eliminating a.b,c from (1), (2), (3), we get |\begin{array}{ccc} 1 & - x & x \\ y & 1 & - y \\ z & - z & - 1 \end{array}| = 0$

$\begin{array}{l} \Rightarrow 1 (- 1 - y z) + x (- y + y z) + x (- y z - z) = 0 \Rightarrow - 1 - y z - x y + x y z - x y z - z x = 0 \\ \Rightarrow x y + y z + z x = - 1 \end{array}$