The plane is parallel to both $\langle -2,-1,-1 \rangle - \langle 0,1,0 \rangle = \langle -2,-2,-1 \rangle$ and $\langle 0,1,1 \rangle - \langle 0,1,0 \rangle = \langle 0,0,1 \rangle$. The plane passes through the point $\langle 0,1,0 \rangle$ so a parametrization for the plane is ${\bf r}(s,t)= \langle 0,1,0 \rangle + s\langle -2,-2,-1 \rangle + t\langle 0,0,1 \rangle$. You can think of this as standing at the point $\langle 0,1,0 \rangle$ and then moving any amount in either $\langle -2,-2,-1 \rangle$ or $\langle 0,0,1 \rangle$ direction to get around on the plane.
To find the scalar equation for the plane you need a point and a normal vector (a vector perpendicular to the plane). You already have a point (in fact you have 3!), so you just need the normal. You've already constructed 2 vectors which are parallel to the plane so computing their cross product will give you a vector perpendicular to the plane.
$$ \langle -2,-2,-1 \rangle \times \langle 0,0,1 \rangle = \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ -2 & -2 & -1 \\ 0 & 0 & 1 \end{vmatrix} = \langle -2,2,0 \rangle$$.
Using the normal vector $\langle -2,2,0 \rangle$ and the point $\langle 0,1,0\rangle$, the scalar equation of the plane is $(-2)(x-0)+(2)(y-1)+(0)(z-0)=0$.
Remember the parametrization is "point plus 2 parameters and 2 parallel vectors" the scalar equation is "point plus 1 vector perpendicular to the plane"
A related problem. If you know the normal vector $n=(n_1,n_2,n_3)$ to a plane and a point $p=(x_0,y_0,z_0)$ lies in the plane, then we can find the equation of the plane as
$$ n.(X-p)=0 \,,$$
where $X=(x,y,z)$ an arbitrary point lies in the plane. The point is not a problem, since you have three of them $p_1=(0,0,0)\,,p_2=(1,2,-1)\,, p_3=(0,1,1)$. The task is how to find the normal vector to the plane. I believe, you have studied the cross product of two vectors and you know the fact that the cross product of two vectors is a vector perpendicular to the plane that contains these two vectors.
Now, since you have three points, you can form two vectors
$$ v_1=p_2-p_1 \,, \quad v_2 = p_3-p_1 \,.$$
Once you form $v_1$ and $v_2$ you can find the normal to the plane as
$$ n = v_1 \times v_2 \,.$$
Now, you should be able to find the equation of the plane $P_1\,.$
Best Answer
Hint: Write the equation as $$ax+by+cz+d = 0,$$ with $(a,b,c)$ being a normal vector to the plane. Then we find $d$ using one point of the plane, for example, $(0,1,1)$. This direction can be found computing the cross-product: $$a{\bf e}_1+b{\bf e}_2+c{\bf e}_3 = \begin{vmatrix} {\bf e}_1 & {\bf e}_2 & {\bf e}_3 \\ 1 & 0 & 1 \\ 2 & 1 & -1\end{vmatrix}.$$