Write a system of three linear equations.with three variables each that has the unique solution $(-1,1,2)$. Write your equations in standard form. Use the variables $x, y$, and $z$ with non-zero coefficients.

Step 1 ：First, we know that the solution to the system of equations is $(-1,1,2)$. So, we can create three equations using these values for $x, y$, and $z$ respectively.

Step 2 ：Let's start with the first equation. We can choose arbitrary coefficients for $x, y$, and $z$ as long as they are non-zero. Let's choose $2, 3$, and $4$ respectively. So, the first equation is $2x + 3y + 4z = k$. Substituting $x=-1, y=1$, and $z=2$ into the equation, we get $2(-1) + 3(1) + 4(2) = k$. Simplifying, we get $k = 5$. So, the first equation is $2x + 3y + 4z = 5$.

Step 3 ：For the second equation, let's choose different coefficients, say $5, 6$, and $7$. So, the second equation is $5x + 6y + 7z = k$. Substituting $x=-1, y=1$, and $z=2$ into the equation, we get $5(-1) + 6(1) + 7(2) = k$. Simplifying, we get $k = 9$. So, the second equation is $5x + 6y + 7z = 9$.

Step 4 ：For the third equation, let's choose yet different coefficients, say $8, 9$, and $10$. So, the third equation is $8x + 9y + 10z = k$. Substituting $x=-1, y=1$, and $z=2$ into the equation, we get $8(-1) + 9(1) + 10(2) = k$. Simplifying, we get $k = 11$. So, the third equation is $8x + 9y + 10z = 11$.

Step 5 ：So, the system of equations that has the unique solution $(-1,1,2)$ is: $\boxed{2x + 3y + 4z = 5}$, $\boxed{5x + 6y + 7z = 9}$, and $\boxed{8x + 9y + 10z = 11}$.