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
$$(\vec r-\vec r_M)\cdot \vec N=0$$
$\vec r_P=\hat i 7+\hat j 7-\hat i 5$
$\vec r_Q=-\hat i 5+\hat j 1-\hat i 1$
$\vec r_M=\frac12 (\vec r_P+\vec r_Q)=\hat i 1+\hat j 4-\hat i 2$
$\vec N=\vec r_Q-\vec r_P=-\hat i 12-\hat j 6+\hat i 6$
$\vec N\cdot (\vec r-\vec r_M)=(-12 x-6y+6z)-(-12-24-12)=0\Rightarrow 12x+6y-6z=48$
Say that you have
a plane: $ax+by+cz=d$
and a point: $(f,g,h)$.
The basic idea is that because of the ways that planes are constructed (with two different vectors), a perpendicular one will be the cross product of these vectors. Consequently, you can always find a perpendicular vector to a plane (think about putting a stick in the mud, and moving it around a 2-d surface.) This vector looks like: $[a,b,c]$.
For a "proof" of this, consider the following 2 points on the plane:
$(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$.
Since they are on the plane, they both satisfy the following equation:
$$ax_{1}+by_{1}+cz_{1}=d$$ and $$ax_{2}+by_{2}+cz_{2}=d$$
Now, by subtraction we have that: $$(ax_{1}-ax_{2})+by_{1}-by_{2}+cz_{1}-cz_{2}=0$$
or, better yet:
$$[a,b,c]\cdot[x_{1}-x_{2},y_{1}-y_{2},z_{1}-z_{2}]=0$$
Since $[x_{1}-x_{2},y_{1}-y_{2},z_{1}-z_{2}]$ is a vector made by points on the plane, it is a vector on the plane. Since the dot product is zero, $[a,b,c]$ is perpendicular to this vector.
Now you more explicitly want a "line" given by a parametric that has the direction of $[a,b,c]$ and that goes through the point $(f,g,h)$. But then, just look at: $(x,y,z)=(f,g,h)+t[a,b,c]$. Then you have three different equations for $x,y,z$. Substitute them into the original equation for the plane:
$(x,y,z)=(f,g,h)+t[a,b,c]$ implies that:
$x=f+ta$
$y=g+bt$
$z=h+ct$
Then, by substitution into the first equation for the plane:
$a(f+ta)+b(g+bt)+c(h+tc)=d$
Now you have one variable ($t$) and a bunch of constants so you can solve for $t$ and figure it out!
To provide further intuition: take the stick in the mud picture, and imagine, say... some piece of dust floating (and static) in space above the ground. First you want to get the stick directly under the piece of dust (orient it in the mud so that it is perpendicular and right below the piece of dust)-- then solve for how long the stick needs to be until it reaches it, (solve for the scalar, $t$.) If you can do that, you will find a "line" that goes through the point and intersects the plane perpendicularly.
Best Answer
Assuming the incident ray is coming from the negative side of the plane $x=0$.
Parametric equation of incident ray is $$\mathbf{r}(t)=(0,2,1)+t(2,2,0)$$ where $t\in (-\infty,0]$.
The incident ray passes through $(0,2,1)$ which is on the reflecting plane. That's the point of incidence.
The reflected ray is just simply reverse the $x$-component of the tangent vector.
So the reflected ray is $$\mathbf{r}(t)=(0,2,1)+t(-2,2,0)$$ where $t\in [0,\infty)$.