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 $P_1\,.$
Your calculation of the cross product is incorrect. You should have $n_1\times n_2 = (-14, 7, 7)$. I imagine, once you fix that, you should have the plane you desire as you are using the correct method.
Best Answer
If you're using the cross product, this gives $$ \begin{align} \begin{vmatrix} e_{\vec{x}} & e_{\vec{y}} & e_{\vec{z}} \\ 1 & 0 & \sqrt{3} \\ 1 & \sqrt{3} & 0 \end{vmatrix} = \left(\begin{vmatrix} 0 & \sqrt{3} \\ \sqrt{3} & 0\end{vmatrix}, -\begin{vmatrix} 1 & \sqrt{3} \\ 1 & 0\end{vmatrix}, \begin{vmatrix} 1 & 0 \\ 1 & \sqrt{3}\end{vmatrix}\right) = \left(-3, \sqrt{3}, \sqrt{3} \right) \end{align} $$
Thus, $\vec{p} = t(-3,\sqrt{3},\sqrt{3}), \enspace t\neq 0$ is a orthogonal vector to vector $\vec{u}$ and $\vec{v}$ and the plane they make. If $(x,y,z) and (x_0,y_0,z_0)$ are points on the plane, then $\vec{w} = (x-x_0, y-y_0,z-z_0)$ is a vector on the plane, and $\vec{w}\cdot\vec{p} = 0$:
$$ \begin{align} t(-3,\sqrt{3},\sqrt{3})\cdot(x-x_0, y-y_0,z-z_0) &= 0 \\ t(-3(x-x_0)+\sqrt{3}(y-y_0)+\sqrt{3}(z-z_0) &= 0 \end{align} $$
We can choose $t=1$ (because a vector is orthogonal regardless of its length) and $(x_0,y_0,z_0) = (0,0,0)$ (because the plane passes through the origin, because both vectors go through the origin). This yields:
$$-3x+\sqrt{3}y+\sqrt{3}z = 0$$