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 $P_1\,.$
Let $\vec r_P$ and $\vec r_Q$ represent two vectors.
The vector from $P$ to $Q$ is $\vec N=\vec r_Q-\vec r_P$, which is normal to a plane.
The midpoint $\vec r_M$ between $P$ and $Q$ is $\vec r_M=\frac12 (\vec r_Q+\vec r_P)$.
The plane that passes through $\vec r_M$ and is normal to $\vec N$ can be expressed by
$$(\vec r-\vec r_M)\cdot \vec N=0$$
$\vec r_P=\hat i 7+\hat j 7-\hat i 5$
$\vec r_Q=-\hat i 5+\hat j 1-\hat i 1$
$\vec r_M=\frac12 (\vec r_P+\vec r_Q)=\hat i 1+\hat j 4-\hat i 2$
$\vec N=\vec r_Q-\vec r_P=-\hat i 12-\hat j 6+\hat i 6$
$\vec N\cdot (\vec r-\vec r_M)=(-12 x-6y+6z)-(-12-24-12)=0\Rightarrow 12x+6y-6z=48$
Best Answer
You have the normal to the plane $(2, -1, 2)$. This is the direction vector of your line. The line passes through $(1,2,1)$ so the line equation is $$ (x,y,z)=(1,2,1)+\lambda(2, -1, 2) $$ where $\lambda$ is the parameter. Then you have 3 individual equations, one each for $x,y$ and $z$ in terms of your parameter.
For example, $x=1+2\lambda$.