Question : If $x+y+z=0$, then what is the value of $\frac{\left (3y^{2}+x^{2}+z^{2} \right )}{\left (2y^{2}-xz \right)}$?
Option 1: $2$
Option 2: $1$
Option 3: $\frac{3}{2}$
Option 4: $\frac{5}{3}$
Correct Answer: $2$
Solution :
Given: $x+y+z=0$
Solution:
$\frac{(3y^{2}+x^{2}+z^{2})}{(2y^{2}-xz)}$
Since $x+y+z=0$, put $x = 1$, $y = –1$ , $z = 0$;
$=\frac{3+1}{2}= 2$
Hence, the correct answer is $2$.
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