Given:A plane passing through the points $A\left(x_1,y_1,z_1\right)\equiv \left(1,1,-1\right)$, $B\left(x_2,y_2,z_2\right)\equiv \left(2,-1,0\right)$ and $C\left(x_3,y_3,z_3\right)\equiv \left(-1,0,2\right)$. 

$\therefore$Equation of plane passing through $A,B \text{\&} C$ is

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-1& y-1& z+1\\ 2-1& -1-1& 0+1\\ -1-1& 0-1& 2+1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-1& y-1& z+1\\ 1& -2& 1\\ -2& -1& 3\end{array}\right|=0$

$\Rightarrow \left(x-1\right)\left(-6+1\right)-\left(y-1\right)\left(3+2\right)+\left(z+1\right)\left(-1-4\right)=0$

$\Rightarrow -5x-5y-5z+5=0$

$\Rightarrow -x-y-z+1=0.......\left(1\right)$

Putting $x=1, y=2, z=-2$

$-1-2+2+1=0$

$0=0$

Hence, points $\left(1,2,-2\right)$ satisfy the equation of plane so $\left(1,2,-2\right)$ lies on the plane $\left(1\right)$.

Putting $x=2,y=1,z=-3$ in $\left(1\right)$, we get

$-2-1+3+1\ne 0$

$\therefore$Point $\left(2,1,-3\right)$ does not lie on the plane $\left(1\right)$.

Again putting $x=2,y=-2,z=2$ in $\left(1\right)$, we get

$-2+2-2+1\ne 0$

$\therefore$Point $\left(2,-2,2\right)$ does not lie on the plane $\left(1\right)$.

Also putting $x=2,y=1,z=2$ in $\left(1\right)$, we get

$-2-1-2+1\ne 0$

$\therefore$Point $\left(2,1,2\right)$ does not lie on the plane $\left(1\right)$.

Hence, option $1$ is correct.