A system is given by x+y+z=6, 2x-y+z=3, and x+2y-z=2. What is the value of z?

Add equations 1 and 2: $3x+2z=9$. Add equations 1 and 3: $2x+3y=8$. To get a third relation, subtract equation 1 from 2: $x-2y=-3$, so $x=2y-3$. Substituting into $2x+3y=8$ gives $4y-6+3y=8$, so $y=2$ and $x=1$. Then $z=6-1-2=3$, choice C. Check in equation 3: $1+4-3=2$, which is correct. Choice B ($2$) is the value of $y$, and choice A ($1$) is the value of $x$. Choice D ($4$) does not satisfy equation 1 when paired with any consistent $x$ and $y$ from the system. Always reread which variable the question asks for.