Fiest be aware that you can define a plan by different means.
One of them is through parametric equations such as
$(1)$ $x=a_1+\lambda u_1 +\mu v_1$, $y=a_2+\lambda u_2 +\mu v_2$ and $y=a_3+\lambda u_3 +\mu v_3$
with $(a_1,a_2,a_3)$ a point of the plane and $(u_1,u_2,u_3)$, $(v_1,v_2,v_3)$ two non colinear vectors.
An other one is through a cartesian equation such as
For your question $(a)$, your professor gave you a parametric equation, which is different from the statement of your problem (give the cartesian equation).
In any case, let's find a parametric equation for the problem $(a)$.
Since you already have a point of the plane $(2,1,-2)$, you just need to find two non-colinear vectors of the plane. Each of them should be perpendicular to the normal vector, that is, the dot product should be $0$.
One of the vector could be very easily $(1,1,0)$, whom dot product with $(1,-1,2)$ is clearly $0$.
The second one could be $(0,2,1)$. Since it is not colinear with $(1,1,0)$, this works.
Hence an equation $(2,1,-2)+\lambda (1,1,0)+ \mu (0,2,1)$
This is different from the answer of your professor, but this is OK, there are an infinity of possible parametric equations for the same plan. Bear in mind that a method to find the last vector, instead of finding our one fitting one, could be to find a vector which is normal to both the vector normal to the plane and the one you already found, with the vector cross product. It would be in that case $(1,-1,2) \times (1,1,0)$. It is probably what your professor did.
In the second case, you can take the first point $A$ as your defining point, and the vectors $AB$ and $AC$ as vectors for the parametrization of the plane.
Or you can also (for cartesian equation) find out the three unknowns by resolving this set of equations:
$ax_1+by_1+cz_1+1=0$, $ax_2+by_2+cz_2+1=0$, $ax_3+by_3+cz_3+1=0$