So, I took a slightly different approach.
I multiplied the top equation by two and added to get
What we want is two points, much more than a vector (right now). So I'll arbitrarily choose and . Those clearly satisfy (1) and if you substitute, you get the points:
The reason you want two points is because for (a) now we have three points, which easily produce two vectors.
So the vector you pointed out: , is from (1,2,1) to (-1,-11,-6). The other one we'll use is from (1,2,1) to (3,1,3) which is . Then we take the cross product.
x (or maybe negative that; as we both know it is irrelevant here). Barring any arithmetic errors, the equation for the plane is
So hopefully that made sense.
As for part (b), you have that vector from before, and you also know the given plane's vector. If you draw this out, you'll see the cross product of these two lie inside the plane that we want, and in fact form the best second vector for cross product with . With less English, we want:
Now we just use one of those two magic points:
. Divide by -222:
So, final answer: