The comments give one way of understanding, but I figured I could maybe explain the original proof better.
Suppose our plane has normal vector $n$ (which is normal to the plane at every point on it) and an example point on the plane: $p$. These two vectors, one a direction and the other a point, determine any plane.
Then, our plane equation is $$ n\cdot(x-p)=0 $$
that is, every $x$ that satisfies this equation is a member of the plane. This is equivalent to $n\cdot x = n\cdot p$.
Notice that $x=p$ is a member of the plane. But, $x=p+n$ is not a member, nor is $x=\vec{0}$, assuming $p\ne \vec{0}$.
Another way to write this is as a point set $\Pi$:
$$
\Pi = \{ \;x\in\mathbb{R}^3\;|\;n\cdot x= n\cdot p\;\}
$$
i.e. these are the set of points making up the plane. Again, notice that if $p=\vec{0}$, then the origin is a point in $\Pi$. But if $p$ does not vanish, then the origin is not in the plane.
The reason is that $p$ is a translation or shifting parameter. That is, a plane has an orientation parameter, $n$, which "rotates" it, and a position parameter $p$, which slides the plane around. When $p=\vec{0}$, we have slid the plane so that it intersects that origin. The plane equation in this case is $n\cdot x = 0$.
Here's a different approach. Every plane is determined by giving 3 unique points. Let's take $p,a,b$. Define $T_1=a-p$ and $T_2=b-p$. We can suppose $T_1$ and $T_2$ are orthogonal; if they are not, we can use Gram-Schmidt orthonormalization.
Now, suppose we walk around between $p$ to $a$ or $b$. This is the same as adding some multiple of $T_1$ or $T_2$ to $p$. So a parametric equation for the plane is
$$
x(s,t) = p + sT_1 + tT_2
$$
so that if you input any $s$ and $t$, your output is a point on the plane.
See also here.
Notice that a normal vector is simply $n=(T_1\times T_2)/||T_1\times T_2||_2$.
So we get equivalence to $n\cdot(x-p)=0$ as before.
Let $\mathbf{OP} = <a,b,c> $.
We know that $AB$ and $\mathbf{OP}$ must be perpendicular, so the dot product of $\mathbf{OP}$ and $<2,2,-2>$ is $0$ ($<2,2,-2>$ plays the role of $\mathbf{v}$ in the diagram below)
$$<2,2,-2> \bullet <a,b,c> = 2a+2b-2c = 0$$
$$ \Rightarrow a+b-c=0$$
We also have that for some value of $\lambda$ (let's just call it $t$)
$$<1,2,2> + <2,2,-2>t $$
$$=<1+2t, 2+2t, 2-2t> = <a,b,c>$$
because $P$ lies in the $AB $
Combining the two equations, we get
$$1+2t +2 +2t - 2 +2t = 0 \Rightarrow 6t = -1$$
$$\Rightarrow t = -1/6$$
So $\mathbf{OP} = <2/3, 5/3, 7/3> = \frac13 <2,5,7>$
Best Answer
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
$$(\vec r-\vec r_M)\cdot \vec N=0$$
$\vec r_P=\hat i 7+\hat j 7-\hat i 5$
$\vec r_Q=-\hat i 5+\hat j 1-\hat i 1$
$\vec r_M=\frac12 (\vec r_P+\vec r_Q)=\hat i 1+\hat j 4-\hat i 2$
$\vec N=\vec r_Q-\vec r_P=-\hat i 12-\hat j 6+\hat i 6$
$\vec N\cdot (\vec r-\vec r_M)=(-12 x-6y+6z)-(-12-24-12)=0\Rightarrow 12x+6y-6z=48$