Question : If $\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}$ and $x+y+z\neq 0$, then each ratio is:
Option 1: $\frac{1}{a-b-c}$
Option 2: $\frac{1}{a+b-c}$
Option 3: $\frac{1}{a-b+c}$
Option 4: $\frac{1}{a+b+c}$
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Correct Answer: $\frac{1}{a+b+c}$
Solution :
Given: $\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}$
Let $\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}$ = k
Or, $\frac{x+y+z}{xa+yb+zc+ya+zb+xc+za+xb+yc}$ = k
Or, $\frac{x+y+z}{y(a+b+c)+x(a+b+c)+z(a+b+c)}$ = k
Or, $\frac{x+y+z}{(a+b+c)(x+y+z)}$ = k
$\therefore$ k = $\frac{1}{(a+b+c)}$
Hence, the correct answer is $\frac{1}{(a+b+c)}$.
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